SUPPLEMENT- TO THE THENS AN EXPERIMENT USING PROGRAMED MATERIAL?" IN TEACNING A NONCREDET ALGEERA COURSE; 5'51“: 5???? 4i AT THE COLLEGE LEVEL Thom form» 909m: m. p , MICH§GAN STATE UNIVERSITY Elame Vman Akron 1965 v‘ f! 1.33;»; ETW‘TT‘T” ~~ - 7,: LIBRARY lllllHijlHIVIIHIIHWNHIIUIHIIHIHlllllllllllllllllll MichiganState "' IIIIIIIIIIIIIIIIIIIIIIIIIIIIIII 321 93 00621 8808 Umvcrsxty PLACE IN RETURN BOX to remove this checkout from your record. TO AVOID FINES return on or before date due. DATE DUE DATE DUE DATE DUE MSU Is An Affirmative Action/Equal Opportunity Institution SUPPLEMENT TO THE THESIS AN EXPERIMENT USING PROGRAMED MATERIAL IN TEACHING A NONCREDIT ALGEBRA COURSE AT THE COLLEGE LEVEL By Elaine Vivian Alton A THESIS Submitted to Michigan State University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY College of Education 1965 Copyright by ELAINE VIVIAN ALTON .19 E E Preface Increasing enrollments have brought to colleges a number of students who are inadequately prepared to complete college algebra successfully. These students need to relearn many of the basic ideas of elementary algebra. This material was written for the student who needs to relearn and review these basic ideas. 'While most students will have some previous background in algebra, it is assumed only that students are able to do addition. subtraction, multiplication and division with positive integers. Therefore a review of arithmetic as well as a complete treatment of the basic ideas of algebra through quadratic equations are included. Thus this material may be used by students who have no previous background in algebra. The answers to the problems are more complete than those usually fbund in a textbook or workbook. This is to give a student additional help with the intermediate steps of a complex problem. iii TABLE OF CONTENTS PREF‘ACEO o o 0 D I a I n o O O O 0 INSTRUCTIONS o o o o o o o e o o a CHAPTERS l. 0 REVIEW OF ARITHMETIC, SIGNED NUMBERS, . . , . OPERATIONS WITH POLYNOMIALS FACTORING . o o O O I O o I FRACTIONS , . LINEAR AND FRACTIONAL EQUATIONS EXPONENTS AND RADICALS QUADRATIC EQUATIONS , , Page iii N 69 104 144 174 207 234 252 254 256 25? Instructions A series of questions and incomplete statements appear on the following pages. You are to answer the questions or fill in the blank or blanks. The information needed is contained in the statement or in previous statements. Read each frame completely before writing your answer. You must read carefully to make sure your answer is the correct one. Clues such as underlined words and the first letter of missing words are often given. Pay attention to these. Compare your answer to the correct one. The correct one is given to the left of the next statement. It is recommended that you use an index card or a piece of paper to hide the answer until you have decided what the answer should be. Ybu will probably learn more as reading an answer is not the same as constructing one yourself. The eye is quicker than the will power of the most honest. If you make an error. make sure that the given answer makes sense to you before proceeding to the next frame. For continuity of thought. take a few minutes at the beginning of each work session to reread a few frames immediately preceding your last answer. Chapter 1 - Review of Arithmetic The algebra presented in this book deals with the applications of the processes of arithmetic to symbols and letters which represent numbers. In order to do algebra, you will need an understanding of the arithmetic processes. YOu will also need to do these processes with a certain amount of Speed. The only way to obtain understanding. accuracy and Speed is through practice. Therefore. it is necessary to review the arithmetic processes. 1. 14 positive integers 2. 2. an infinite number 3. 3. No. 0 is an integer but is 4. go_t a positive integer. 4. Yes. (See answer to frame 3.) 5. 5. subtraction. multiplication. 6. and division. 6. tenm 7. The numbers you first dealt with in arithmetic were the numbers 1. 2. 3, ... These are called the counting numbers or the positive integers. 2. #23 and 17 are examples of How many positive integers are there? Is 0 a positive integer? Is 0 an integer? YOu dealt with four fundamental Operations in arithmetic. These Operations are addition, . . and . We deal with these same Operations in algebra. These operations obey certain basic laws and we shall find out about these laws as well as reviewing arithmetic techniques. In addition. the numbers to be added are called terms. In 2+5+7. the 5 is called a . The result of addition is called the E921. 2+5 can be read "find the of 2 and 5". 9. 10. ll. 12. 13. sum 4+3 or 11-4 sum two two terms terms are (3—1) and 5 (these can be given in any order) multiplied by 01‘ times - 3 - 8. 9. IO. 12. 13. IN. 2+5 also represents the sum of 2 and 5. Most of you would probably say that 7 was the sum of 2 and 5. However. 7 is another numeral which represents the same number as 2+5 represents. Using 4 and one other number. write another numeral which represents the same number as 2+5 represents. A+3 represents the of 3 and h. Write a single numberal which represents the sum of 3 and h. We Often use () parentheses. brackets[]. and braces to indicate that a group of terms is to be treated as a single number. Fer example. 5+(3-l) indicates that we are to add (how many) numbers? In 5+(3-l). how many terms are there? What are the tenms? It is customary in arithmetic to use the symbol "x" to mean multiplication. It is more convenient in algebra to use either a dot - or () to indicate multi- plication. Instead of writing 2 x 3. we can write 2'3. 2(3) or (2)(3). All of these mean 2 3. The result of multiplication is called the product. ‘What is the product of 7 and 9? 14. 15. 16. 17. 18. 19. 20. 21. 22. - A - ?“9 or 7(9) or (7)(9) or 63 15. “'59 “(5): 00(5): 20 16. any three of the above 4'3. or 4(3) or (3)(#) 17. the 4 and the 3 can be listed in any order. 2 and 7 18; or 1 and 14 three factors 19. the factors are 2. (h—l) and 3 No because factors are 20. associated with products and this is a sum. or No as no multiplication is indicated. a term 21. (Refer to frame 6 if necessary) two terms 22. the terms are 3-5 and 2 No. The 5 is a factor of the 23. second term. It is n_p__t a factor of the whole eXpression as it is £125 multiplied by 24-3. Most of you probably wrote 63 as the product of 7 and 9. Note that 7-9. 7(9) and (7)(9) are different symbols which represent the product of 7 and 9. 63 is another symbol which also represents this product. write the product of A and 5 in three different ways. Express 12 as the product of two numbers each of which is less than 5. In 3'4, the 4 and the 3 are known as factors. Factors are numbers which are multiplied together or are divisors of the product. What are the factors of IA? wa many factors are there in 302(4-1)? List the factors. In 2+3. is the 2 a factor? Give a reason for your answer. What is the 2 in 2+3 called? In 2+3-5. there are how many terms? What are they? In 2+3-5. is the 5 a factor of the whole quantity? Give a reason for your answer. In (2+3)5a is the 5 a factor of the whole quantity? 23. 24. 25. 26. 27. Yes because it is multiplied by 2+3 or because it is a divisor of (2+3)5. 5(1+6) commutative 3(5) = 5(3) only the order of the two numbers is changed Given two nonnegative integers to add or'multiplyt the order of the two numbers 1 8 illmlaterial e - 5 - 24. 25. 26. 27. 28. Note carefully the difference between 2+3-5 and (2+3)5. When we talk of factors. we usually mean factors of the complete expression. In A.3+5(1+6)+13(9-2)(11+7). the second term is . If we add or multiply any two positive integers together. it is evident that the order of the numbers is immaterial. This is known as the gommutative law. FOr example. h+7=7+fi illustrates the law for addition. Complete the following statement to show the commutative law of multiplication. 3(5) = We postulate the commutative law when the numbers are positive integers. After the operations of addition and multiplication are defined for negative integers. rational numbers and irrational numbers. we can prove the commuta- tive law true for all real numbers. FOr now. we will only deal with the commutative law for positive integers. Commutative Law If a and b are positive integers. then a+b=b+a and aob=b-a. State the commutative law in words. 4'+ (5-2) indicates that numbers are to be (how-many) added? - 5 - 28. two 29. (5-2) + a Remember change 5391 the order of twp numbers. 29. (“+6)3 or (4+6)(3) 30. Thu must use () or brackets [] or braces around the 4+6 to show thatiyou are considering this as ggg,number. 300 faCtor 31. (see frame 17 if necessary) 31. Add 7 and 8 and then add this 32. sum to 2. By the commutative law of addition 44- (5-2) = By the commutative law of multiplication 3(A+6) = (n+6) in 3(4+6) is called a NO matter how many numbers you are given to add or multiply. only 2 numbers are added or multiplied at one time. We need to know what it means when there are more than 2 numbers to be added or multiplied. Consider 2+7+8. 'We could add the 2 and the 7 and then add the result to 8. Using parentheses to show this: 2+7+8 = (2+7)+8. Without changing the order of the numbers. how else could you find the value of 2+7+87 Expressed using parentheses. we have 2+7+8 = 2+(7+8). In other words. no matter hOW'We group the numbers in addition. the sum is the same. This is called the associative law and the associative law applies to both addition and multiplication. The associative law is postulated true for'positive integers. Associative Law If a. b and c are positive integers. then a+b+c = (a+b)+c = a+(b+c) and aoboc = (a-b)c = a(b-c). 320 33. 35. 2(5-7) 33. Note that we change only the way the numbers are grouped together net,the order of the numbers. (h+5)+3 = 4+(5+3) 3A. in either order (a) commutative law for 35. addition (b) associative law for addition (c) associative law for multiplication (d) commutative law for multiplication 6) commutative law for multiplication (2-5)8 = 8(2.5) by the 36. commutative law of multi- Plication By the associative law of multiplication. 2-5-7 = (2-5)? = By the associative law of addition. use = State whether the following are true because of the commutative law or the associative law. Specify the operation (addition or multiplication). (a) 3+(6+4)=(6+h)+3 (b) 7+(8+#)=(7+8)+4 (c) (3°6)-5= 3(6°5) (d) 2(7+9)=(7+9>-2 (e) 2(3~5)=(3-5)(2) To show that 5+(6+4) = (5+4)+6. we need to use both the commutative and associative laws of addition. 5+(6+4) = 5+(4+6) by the commutative law of addition 5+(4+6) = (5+4)+6 by the associative law of addition Therefore. 5+(6+h) = (5+#)+6 Show that (2o5)8 = (8-2)5. State reasons. Show that (3o6)9 = 9(6-3). State reasons. 8(2-5) = (8.2)5 by the associative law of multiplication. Therefore. (2-5)8 = (8°2)5. 36. 38. 39. 40. 41, (3-6)9 = 9(3-6) by the comm. law of multiplication. 9(3-6) = 9(6-3) by the comm. law of multiplication. Therefore. (3-6)9 = 9(6-3). 3°5+3°4 distributive commutative (748)13 by the comm. law = 13.7+13.8 or 7ol3+8-13 by the distributive law. use) 37. 39. 40. 41. 42. -8- The distributive law connects the operations of addition and multi- plication. Distributive Law If a. b and c are positive integers. then a(b+c) = a-b + a-c. For example. 2(7+3) = 2.7+2.3. By the distributive law. 3(5+4) = . 8(6+9) = 8o6+8°9 because of the law. 8(6+9) = (6+9)8 by the law of multiplication. Thus both 8(6+9) and (6+9)8 are equal to 8°64-8'9. Using the commutative law of multiplication first and then the distributive law. complete 13(7+8) = _. = An equality is true whether read from left to right or from right to left. Thus the distributive law not only states that a(b+c) = a°b+a0c but also that a°b+aoc = a(b+c). Hence. 8°5+8°4 = Complete using the distributive law. (a) 5(l+3) = (b) 5-2+5°3 = (c) 12-4+12-5 : 42. “Be 45. 4?. (a) 5'1+5°3 (b) 5(2+3) (0) 12044-5) N) 44152 = nose-5 NOte - if you wrote 4o3+5 on 43. the left side of the equation. it is wrong. multiply add 207 and 2-3 2(743) = 2.10 1%z20 addition 8 and “(54-6) 4o3+5 states that 9931 3 is multiplied by 4. You need the number (3+5) to be multiplied by 4. (e) 3°3i'2 = 54.21;). 20 45. (d) 4' ._____ '3+ ________ ‘5 (6) 3° __ + __ '2 = 5. By the distributive law. 2(7+3) = 2.7+203. This means that when 2(7+3) is evaluated we must get the same result as when 2°7+2~3 is evaluated. 2(7+3) means to (what Operation) 2 and (7+3)? 2'7+2°3 means to (what Operation on what numbers) (7+3) is another symbol representing 10. therefore 2(7+3) = 2- .-.- 2.7 is another symbol representing 14. g.3 is another symbol representing Therefore. 2'7+2°3= = In 8+4(5+6). what ggg_operation connects all the numerals? What terms are to be added if all the numerals are to be used? Then we must evaluate 4(5+6) before we can add it to 8. 4(5+6) is another symbol representing 50. 51. 52. 53. 55 - 10 _ 44 49. to evaluate 4(5+6). we can say this is 4'11 which equals 44 or we can use the distributive law so 4(5+6) = 4°5+4°6 and this equals 20+24 which equals 8+44 = 52 multiplication factors (2+4) and (19+2) (6)(21) = 126 2-19+2-2+4o19+4-2 = 38+4476 (3+4)(5+6+2) = 3(546+2)+4(5+6+2 ) = 3'5+3-6+3°2+4 5+4- -6w 15+18+6+20+24+8 = 91 or (3m)(5+6+2 ) = 3' 13 + 4. '13 = 39+52 = 91 50. 51. 52c 53. is 55 Hence. 8+4(5+6) = 8+ = To simplify (2+4)(l9+2). first determine the gag operation which connects all the numbers. This is Numbers which are to be multiplied are called (see frame 17 if necessary) The factors in (2+4)(19+2) are and Thus (2+4)(l9+2) = and this equals (2+4)(19+2) can also be evaluated usin the distributive law. (2+4 (19+2) = 2(19+2)+4(19+2) by the distributive law. Use the distributive law to simplify (3+4)(5+6+2). . Simplify the following. (a) 2+3(4+5) (b) (2+3)(4+5) (c) 2~3+4'5 (d) (2'3+4)'5 -11- 56. (a) 2+3(4+5) = 2+3(9) = 57. Note that in the last frame. the 2+2? = 29 four problems all contained exactly or the same numbers 2. 3. 4 and 5. 2+3(4+5) = 2+3-4+3-5 = However. they were grouped and 2+12+15 = 29 combined in different ways and so none of the results were equal. (b) (2+3)(4+5)= 5(9) = 45 Make sure you understand the differences in these. (2+3)(4+5)= 2(“+5)+3(4+5)= 2 2 9+3 9 = l& 27: 3 is read "3 squared" and means (3)(3) or 3'3. 33 is read "3 cubed" and means (c) 2'3+4'5 = 6+20 = 26 (d) (2 3+4)50= (6+4)5 = 10 5 = or that there are (6+4)5 = 6’5+4°5 = 30+20 = factors of 3. 50 ‘ (how many) 57. 3(3)(3) 58. 3“ is read "3 raised to the fourth power" and means . three factors 58. 393°3-3 or that there are 59 four factors of 3. 2(2)(2)(2) can be written which is read . 59. 2“ 60. In 24’ the 4 is called an exponent 2 raised to the fourth power and the 2 is called the base. An exponent shows the number of times the base is to be used as a factor providing that the exponent is a positive integer. _ In 46. the 6 is called the . 60 . exponent 61. In 53. the 5 is called the _______. 61. base 62. 23 means and this equals 62, 2'2'2 = 8 63. Evaluate the following. (A)32=_____=__ (b) 25 - 12 - (a) 3'3 = 9 64- (b) 2.2.2.2.2 = 32 w)mmh=eu (d) 5'5 = 25 (e) 3'3'3 = 27 23.52 65. 8°25 = 200 66. 849 = 17 67. exponent 68. (2'5) 69. = 10'10°10‘10 = 10.000 70. - (2+3) used as a factor 71. three times or (2+3)(2+3)(2*3) O . No. (2+3)3 = (2+3)(2+3)(2+3) 23+33 = 2-2-2+3-3-3 and these two quantities dgglt equal the same number. or (2+3)3 means to add 2 and 3 and then cube the result. (e) 43 (a) 52 (e) 33 Write 2(2)(2)(5>(5) exponents. fl using 23.52 23432 In (25)“. the 4 is In (2-5)“, the 1+ indicates that ____________ is to be used as a factor 4 times. Thus <2~5)“ = (2.5)(2.5)(2.5)(2.5). finish evaluating (2-5)”. (2+3)3 means __________. Thus. (2+3>3 = (2+3)(2+3)(2+3> = Is (2+3)3 equal to 23+33? Give a reason for your answer. We shall use the symbol ¥ to mean unequal. So we may write (2+3)3 234-33. 73. 74. 75. 76. 77. - 13 _ 23+33 means to cube 2 and then cube 3 and then to add the results. a! 74. No. 3-2“ = 3-2-2°2-2 and 75. (3°2)4=(3-2)(3-2)(3-2)(3-2) The same factors are not used in both cases. or 3-24 = 3(16) = 1+8 and (3‘2)%= 64 = 1.296 and these aren't equal. (a) 3'16 :2 1+8 76 (b) 4(1) = 4 (c) (4+5)(h+5 = 9'9 = 81 (d) 3-8+9-1o = 24490 -.- 11L» (9) (3-8+9%10 = (24+9>1o = 330 (f) (7)(5) = 7(125) = 875 ( ) 77. (3) 1‘ E3 :2 (e) ¥ Yes (see frame 3 if 78. necessary) Does 3-2“ equal (3~2)“? Give a reason for your answer. Complete the following. (a) 3'42 = ___.____._ = (b) u(1)3 = a (0) (“+5)2 = = (d) 3.23432.1o = = (e) (3'23+32)1o = = (f) (4+3)(1+t+)3 a = . Use either a = or # to make the following true . (a) 15 5 (b) (2+3)2 52 (a) 2'52 50 (d) 73 ______7-13 (e) 4363 3 So far we have dealt with the Operations of addition and multi- plication with positive integers. Let us now consider these two operations and the number 0. Is 0 an integer? O is an integer but is not a positive integer. The positive integers and zero comprise the set of nonnegative integers. -14.. erties of 0 If a is a nonnegative integer, then a+0 = 0 and a'O = O. 78. 156 . 79. 481‘0 = . 79. 0 80. State whether the following are true or false. (a) 16’0 = 16 (b) 16+O = 16 (c) 0+0 = 0 (d) o~o = o 80. (a) false 81. The commutative law holds for the 1 (b) true nonnegative integers. By the L éc) true commutative law. 56+0 = , d)tnm L 81. 04-56 82. By the commutative law 34-0 -.- 82. 0°3h 83. If two nonnegative integers are ‘ multiplied and the product is O, 7 what can.you say about the two 1 numbers? \ 83- One or else both of the 84. This can be stated: 3 nuhbers is 0. If a and b are nonnegative integers. and if a°b = 0, then either a or b or both.must be 0. 0 is known as the identi§x_element of addition because when 0 is added to any nonnegative integer the sum is that same nonnegative integer. 84. Yes. 85. identity element is l. 85. l 86. 86- 2-3 87. $0 this doesn't exist as there 15 29.nonnegative integer Which can.be added to 3 to give 2, - 15 - If there is an identity elanent for’multiplication, it.must be a nonnegative integer such that when it is multiplied by any non— negative integer. the product is that nonnegative integer. In other words. if a is a non- negative integer and if there is an identity element for multi- plication, then a'(the identity element) = a. Is there an identity element for multiplication and if so. what is it? If the product of 3h2 and some number is 342. what is that number? The Operation of subtraction is defined in terms of addition. Subtra on If a and b are nonnegative integers, then a-b is the difference of a and b and a-b = c providing c is a non- negative integer and providing c+b = a. For example. the difference of 6 and 4 is written 6-4 and this exists providing there is a non! negative integer which can be added to 4 and give 6. write the difference of 2 and 3. Does this exist? ‘write the difference of #5 and 27. -16.. 87 o “5'27 88 o 88. (a) 18 because 18+27 = 45 89. (b) 65 because l9+46 = 65 89 o Sllbtrahend 90 o 90. no. 4-7 isn't defined as 91- there is no nonnegative integer which when added to 7 equals 4. 7-4 = 3 because 3+4 = 7 91. no because the difference of 92. and 7 isn't equal to the difference of 7 and 4. Complete the following. (a) 45-27 = because (b) _____r46 = 19 because The difference of 6 and 4 can be written as 6-4. 8 It can also be written as ‘ The number to be subtracted is called the subtrahend. The subtrahend in this case is 4. The 6 in this case is called the mm. In 31-19 = 12. the 19 is called the . If you have been in the habit of choosing the larger of two numbers as the minuend. you must break this habit. The larger number isn't always the minuend. The difference of 4 and 7 is written 4-7. The difference of 7 and 4 is written 7-4. Do both of these represent the same quantity? Give a reason for your answer. Is subtraction commutative? Why? 7-4 can also be read "subtract 4 from 7". - 17 - In order to disprove a state- ment. you only need one counterexample. i. e. , you need only one case that isn't true. 92. 29-14 930 u'8 94. (a) 54-67 this doesn't exist (b) 43(0) or 0(43) this does exist (c) 16-9 this does exist (d) 43+68 or 68+43 this does exist (a) 2-3 this doesn't exist 95o3M91tiplication is commuta- tive, i.e.. when two numbers are multiplied, they can be uSed in any order. 93. 94. 95. 96. ‘Write "subtract 14 from 29". 7-4 also means "take 4 from 7". write "take 8 from 4". 'write symbols expressing the following and tell whether each exists. (a) the difference of 54 and 67. (b) the product of 43 and 0. (c) take 9 from 16. (d) the sum of 42 and 68. (e) subtract 3 from 2. In part b of the last frame why could my answer be written either as 43(0) or as C(43)? Division is defined in terms of multiplication. Dixiaiaa If a, b and c are nonnegative integers and b ¥ 0. then the quotient of a and b (written a§b or 87315 defined as c, (written 3': c) only if c’b = a. For example. «3- = 2 because 2(3) = 6. 96. 3 because 3(13) = 39 97. g is the required quotient 98 This exists if there is a nonnegative integer which when multiplied by 4 equals 0. That is, (some non- negative integer)(4) = 0. 0 is the required number so -18.. 97. 98. the quotient of O and 4 exists and is 0. (a) 6 This equals 14. (b) ;1 This doesn't exist 25 because there is no integer multiplied by 25 which equals 15. 99. Complete: §§ = because We may write the quotient of two numbers, however the quotient doesn't always have to exist. For example, the quotient of 4 and 8 can be written as 4 but 8 this doesn't exist as there is no nonnegative integer which when multiplied by 8 equals 4. Don't forget at this point the only numbers we can use are positive integers and 0. Find the quotient of O and 4. Does this exist? Write the following quotients. State which ones do not exist. For the ones which do exist, find the nonnegative integer which equals the quotient. (a) the quotient of 56 and 4 (b) the quotient of 15 and 25 (c (d) the quotient of 72 and 12 (e (f) the quotient of 5 and O V the quotient of O and 24 V the quotient of 36 and 8 Is division commutative? Give a reason for your answer. - 19 _ (c) __0_ which equals 0. 24 (d) 2; which equals 6. (e) zé'which doesn't exist 8 because there is no integer multiplied by 8 which equals 36. (f) 5 which doesn't exist 0 because every number multiplied by 0 equals 0 or because there isn't any number multiplied by O which equals 5. 99. no. The quotient of 4 and 100. In frame 98. we stated that the 2 doesn't mean the same as quotient of 15 and 25 didn't the quotient of 2 and 4. exist. This is true since we were only concerned with the positive integers and zero. %% represents the quotient of 15 and 25, and this isn't an integer. We shall enlarge our set of numbers so that 15 is defined. 2 5 You dealt in arithmetic with fractions and whole numbers. We are going to call these rational numbers. Definition A rational number is a number of the form % where a and b are integers and b # O. l 5% is now defined to be a number. 100. rational 101. Are the positive integers part of the rational numbers? They are if they can be expressed as the quotient of two integers where the denominator isn't 0. 101. 102. 103. 104. 105. 106. -20.. l 102. yes 103. yes 104. 0 can be expressed as the quotient of 0 and any positive integer. It cannot be expressed as the quotient of 0 and 0 because by the definition in frame 100, the denominator can't be zero. rational 105. no 106. %9 1.5;." or any fraction where 107. the numerator isn't a multiple of the denominator. There are an infinite number of possibilities. The integer 2 can be expressed as 2 divided by what integer? Can every positive integer be expressed as the quotient of itself and one? Can 0 be expressed as the quotient of itself and some other integer? If so. what integer? Then every integer can be expressed as the quotient of two integers where the denominator isn't zero and so every non- negative integer is also a number. Is every rational number also an integer? That is, can the quotient of every two integers be expressed as an integer? Give an example of a rational number which is not an integer. Let us examine %.to see why we must exclude division by O. In any division prdblem such as 6 divided by 2, the quotient g.can be expressed as 3 because 3(2) = 6. 4 So if the quotient 5 exists, then %’must equal some number a such that when a is multiplied by 0 we will get 4. that is 4(0) = 4. ‘We already know that any number multiplied by O is . 107. O 108. is not 109. all except 6 and g 0 O 110. 15, 9 because it is which is another symbol for 6. and O. 1110 10' i 6 another symbol for 0, ;g 4 _ 21 _ 108. 109. 110. 111. Therefore there is g2 number, a, such that E = a because there is ng number, a, such that a(0) = Therefore. % (is. is not) a choose one number. Which of the following represent rational numbers? 0 7. 9. 23 .6. [+9 151 5, Ll" 0, or 0 Which of the rational numbers in frame 109 are also integers? Let us now review the meaning of fractions and the operations on rational numbers. We defined a rational number as a number of the form 3 where a and b b are integers and b # 0. So any rational number can be thought of as the quotient of two numbers. a can also be considered as a.; b b where b # 0. Thus. gg means Since the integers are rational numbers, the Operations on rational numbers are defined so that the basic laws of commutativity. associativity and distribution hold with all rational numbers. Multiplication of rational numbers If a, b. c and d are integers and b a o and d # 0, then - 22 - In other words, multiply the numerators to get the new numerator and multiply the denominators to get the new denominator. Why can't b and d equal 0? 112. division by 0 isn't 113. Then to multiply 5_by'3, we'd defined 7 2 have 5 .3 _ 7 2 - ' 113.1;3=;5_ 114.153.’i=i°1? Why? 7-2 14 2 7 7 2 114. yes because as stated 115. Multiply: (a) _2_4_ . Z in frame 111, the commutative 5 5 law for multiplication is to hold. (b) 3 0 __]_l_ 5 2 (e) 5.‘ lit 3 9 (d) 11 ° 3 ° 5. 7 4 8 (e) 1‘ 11 2 115. ( a) 34-2: ;_6_8_ 116. Euivalent fractions 5'5 25 If a. b and c are rational numbers, and b i O and 0 ¥ 0. (b) 3'11 33 then aa— ac and a/b and ac/bc 3'2 10 b be are called equivalent fractions. (0) 5'14 _ 29. In other words, the numerator 3'9 - 27 and denominator of a fraction may be multiplied by a nonzero 03) remember the associative number and the value of the law applies here fraction will remain unchanged. (J 3) 5 __ 3.8.8 5__ __g£ Change g to an equivalent fraction 2 2 3 having a denominator of 60. (e) ll.can be considered as 11, State the process you used. 1 1°¥=3°J~i=33 2 2 l 2 - 23 - 116. 40 117. 40 is a different rational number €7.66 6'6 both numerator and from 2,. However. it has the denominator were multiplied 3 by 20. same value as 2/3. Note - it is g:gng,to say How many rational numbers are multiply by 20. This there which are equivalent to changes the value and 2/3? dgesn't give an equivalent fraction. 117. an infinite number 118. Change the following to equivalent fractions by filling in the missing parts. i (a) 5. = \ 1+ .53 (b) .5. .-_- .i 3 (e) Z = __ 8 72 (d) 1: _ :2 1+ .— 118 (a) 70 119. Change the following fractions to ‘ (b) 27 equivalent fractions all having } (O) 63 the denominator of 24. l (d) 16 Mg, “ll-2:93 13 1 1 119- 3. = i , _’Z_ = 14 ’ 120. What did you do to determine that I 8 24 12 24 _2 was equivalent to 3 1 a 8 1.1- .66 2“ 8 l g 973 ' 4 ' 24 ~ 4‘1 a 194 ‘ 3 24 - 24 - Z. 120. Multiplied both 121. We can now express __3__ as the numerator and denominator .5 of 3/8 by 3. NOT multiplied 4 by 3. quotient of two integers. we can change to an equivalent fraction by multiplying both numerator and denominator by the same number. If we multiply both numerator and denominator by 12. we will obtain integers in both places. i-i<_12 . g -&(12) v 121. _§’ 122. By using this procedure. it is 1 15 always possible to express the quotient of two rational numbers as the quotient Of two integers or as a rational number. If you are given two fractions to determine whether or not they are equivalent fractions. you can make use of the following. 2. only if b and d aren't d equal to 0 and ad = bc. .E b Using this. determine whether 11/13 and 9/17 are equivalent fractions. 122- no 123. Are 8/12 and 2/3 equivalent 11(17) $ 9(13) fractions? 'Why? 123. yes because 8(3) = 12(2). 124. If we consider this last case, ‘ _§_= 2., you will note that we . 12 3 ‘ have equivalent fractions when ‘ both the numerator and denominator i are divided by the same nonzero number. This comes directly from.our definition of equivalent fractions. 124. (a) 28 Eb; 6/ c 2 (a) 3 (e) 16 (f) 18/4 or 9/2 -25- 1250 The definition stated that .3- and 43% were equivalent fractions if a. b and c were rational numbers and b and c weren't equal to 0. All equalities are true whether read from left to right or from right to left. so g ___. g9 and b be _c_ ____, 3 when b and c don't equal 0. c c Fill in the missing parts so that the following involve equivalent fractions. m (a)Z=__ 8 32 b 16 4 ()24=" (C>i=.. 2 1 (@131- 48 (6)18... 6 48 (f).1.§.-_ 48—12 In the last frame in parts d. e and f we had the same rational number 18 and were to find 48 equivalent fractions or were to find other rational numbers which had the same value as the given one. In part d we obtained 3 ' 8 in part 9 we Obtained 6 . and in 16 125. an infinite number 126. 8 127. 1 WI -26.. 126. 127 . 128. part 1' we Obtained 2L3 . Since 12 all of these are equivalent to 18/48. they must be equivalent to each other. Given a rational number. how many other rational numbers are there which are equivalent to it? 3 is a fraction in its M992 8 terms. A fraction in its lowest terms is one where the numerator and denominator are integers and have no common factors. To reduce a fraction to its lowest terms. divide both numerator and denominator by their common factors. What is the largest common factor of 16 and 24? Then to reduce 1,16; to its lowest 2 terms. we should divide both numerator and denominator by 8. Thus 176; is equivalent to 2, and 2 3 since 2 and 3 have no common factors. 2/ 3 is in its lowest terms. Find a fraction equivalent to i which is in its 108 lowest terms. You may find it difficult to decide if some numbers have common factors or what the largest common factor may be. You don't have to reduce a fraction by using only one step. 128. 111 6 0:5 57 = 3°1 76:2-21 ' 19 or 4 19 or 2 38 - 27 - 129 130. In reducing _36’ you can see that 108 both of these are even numbers and every even number must be divisible by 2. Dividing both numerator and denominator by 2. we obtain 18 We could repeat 51* O the same process as both 18 and 54 are even numbers. Dividing both numerator and denominator by 2, we now get‘Ji . Both of these numbers are 2divisible by 9 and so we get 1/3 which is in its lowest tenms. Reduce 144 120 Express 5% as an equivalent 7 fraction in its lowest terms. It may appear that 57 and 76 have no common factor and in this case express each integer as factors without considering whether they are common factors. Factors of 57 are Factors of 76 are Then 51_ 3012 or 3-12 and in 76 2219 419 either case it is easy to see that both the numerator and denominator have a common factor of 19. or 5% = i , 7 4 Write equivalent fractions for each of the following making sure each result is in its lowest terms. (a) .12 (1972. M198 (d) _é§. 52 24 154 119 - 28 - 130. (a) £31. 131. (b) 3 0r 3. 1 (c):2 7 (d).i 7 131. no 132. 132. term 133. 133. combine the terms in the 134. numerator and denominator 13"- W 1a . a 135. 24 3 p (b) In this problem. the 6 is 1 a factor in both the numerator and denominator. \ so you can immediately : divide both numerator and denominator by the common factors. 10-6 = 1 30°6 3 (c) the 2 1g not a factor here 2 6 1 .. 13 _ 1 i 2 20 -1 ' 39 ’ 3 In 212 is 2 a common factor of 5-2 the numerator and denominator? In the numerator of 2+7. the 2 is called a . Then to write a fraction equivalent to 2:2 what must we 5-2 do before we can decide whether the numerator and denominator have a common factor? 2:2_ 2 and this can be reduced 5-2_3 to 30r3. 1 Reduce the following . (a) 10+6 (b) 10°6 (c) 6 1 30-6 30°6 2 20 -1 (d) 22. (a) 3:11.. 48 4 3 -2 Addition 0 rational numbe s Ifb¥0andifa.bandcare rational numbers. than a. 2.: 1: if _ l. - are b*b ba*bc“meL b In this definition. note that both fractions have the same denominator. Since a is defined to mean 1/b b multiplied by a. we can use the distributive law to find the sum. Find the some of 5/3. 2/3 and 17/3. (d) 1 2 (e) the 3 1g 391 a factor here. 7391.10“ 4 3)-2 10 135. % (5+2+17) = 34 = 8 K») 136. (a) % (l4+7+11) = 3% b l = 29 = 4 ( ) 5 (744+9) 5 (c) Note that this is multiplication and not addition. 11 4 - 29 _ 136. 137. Answers are uSually expressed as equivalent rational numbers where the fractions are in their lowest terms. Ev31uate the following. (a 13 Z 11 9"9‘ 9 (b) Z i 2 5"5 5 (0) 11 ' 1 2 2 According to this definition. we can't add rational numbers unless the denominators are the same. To add 3/4 and 5/7. we must first change each of them to equivalent fractions having the same denominator. Then we can add them by the above definition. When we change fractions to equivalent fractions having the same denominator, we generally find the least common denominator or the smallest number which is evenly divisible by each of the original denominators. We could use any number which is evenly divisible by the original denominators as a common denominator but this would give us larger numbers to work with. The least common denominator for 3/4 and 4/7 is - 3o - 137. 28 138. Change the fractions 3/4 and 5/7 to equivalent fractions and add. 3 .5. .. = a * 7 ‘ 138. 2;. _2_Q_ _ fig; 139. Remember to reduce the final 28 + 28 “ 28 answer to its lowest tenms. Complete the following. (a) .2 . a 12 3 (b) .7. .. 18 * 53'" (c) ll. + .7. _ 2h 18 “ 139. (a) _Z 4‘ 1.6 = Q 140. You may have had some difficulty 12 12 12 in finding the least common (b) 14 + _j__ %E_ denominator (LCD) in part c. 56’ 36 — So far you have obtained the LCD (c) The least common by trial and error. We can use denominator here is 72. the following definition to find 33 'gg _ Q; the LCD more easily. However. 72 * 72 “ 72 first we must know what prime factors are. Pzime numbers are numbers whose only integer factors are itself and 1. For example 6 is n_ot a prime number as the factors of 6 are 3 and 2 or are 6 and 1. The only integer factors of 6 are not itself and 1. 2 is a prime number because the only integer factors of 2 are itself and 1. List the prime numbers that are feund between 2 and 35 inclusive. 140. 2. 3. 5, 7, ll. 13. 17. lbl. What are the prime factors of 19. 23. 29. 31 (a) 1&2 (b) 63 (C) 56 141. (a) 2. 3 and 7 (b) 3: 7 and 3 (c) 2. 2. 7 and 2 142. 10 15 143. factors of 2. 5 and 3 144. no 145, prime factors of 6 are 2 and 3 prime factors of 75 are 50 33116.5 - 31 - 142. 143. 144. 145. You may have listed the prime factors of 42 as 3. 7 and 2. Even though you have listed them in a different order. they are the same factors. A theorem states that there is only one set of prime factors for any integer. Let us now consider finding the LCD for fractions with denominators of 10 and 15. First determine the prime factors of the given denominators. The LCD is composed of every different prime factor in each of the original denominators and each factor appears the maximum number of times it appears in any single denominator. Thus to find the LCD for fractions with denominators of 10 and 15, we use every different factor as many times as it appears in any single denominator. What factors would appear in the LCD if 10 and 15 were the original denominators? Do any of these factors appear more than once in any one of the original denominators? Then the LCD for fractions with denominators of 10 and 15 is 2'3°5 or is 30. Find the LCD for fractions with denominators of 6 and 75. 146. What is the LCD for fractions with denominators of 45. 50 and 6? 146. 147. LCD has factors of 2. 3. 5 It has two factors of 5 as 5 is a factor twice and 5. in one of the original denominators. LCD 2 2°3'5'5 prime factors of 6 are 2 and 3 prime factors of 50 are 595311dz prime factors of 45 are 3.5311(33 LCD = 2°3'5‘5'3 (a) LCD is 3-3'2°2 or 36 511mm: 32: 3.; 12 (b) LCD is 3-5-13 or 195 _‘LQ+.12.=_2&=_& 195 195 195 15 (0) LCD is 36. 144 1 l _ 1 'm*i*£“% (d) LCD is 3'4°7 or 84 2.n=m 84 84 84 (e) LCD is 36 3%*%%*%%=5%% - 32 - 147. 148. ...—‘13 9 Find the following sums. ()1 a9+fi+fi (b).li .li 39"65 ( ) 4 l c # l§'+ 32 (d) 13.+ ll. 21 28 ()4 2.8. e -£'+ 3" 9 Let us consider part b of the last frame again. In order to get the LCD. we found the prime factors and then used each different one the maximum number of times it appears in any single denominator. Our LCD this time was 3°5°l3 or 195. Suppose we hadn't multiplied our factors together but had used 3'5'13 as our LCD] .5. 4 .39... .12.. .52.. 39 8'5 3°5°13 3-5-13 3-5-13 and now we could reduce this to its lowest terms by finding out which of the factors of the denominator are also factors of the numerator. 13 is a factor of both.and dividing both numerator and denominator by this we get .11.. or 4/15. 3'5 -33... 148. LCD is 3°5°2’7 149. M + L + __3...L= 3050207 30502.7 30502.7 140-35-15 = ggo = _;2,= 3050207 3050207 307 12 21 149. (a) LCD is 6. 5.- 5.3.: 3=2§= (b) LCD is 2 3 70 or 42 .2._ .3.= _&.= .2 42 42 u2 21 (c) LCD is 12. 12...;Z_ .41 12 12 12 boll-J 150. Find the LCD of the following fractions and add them keeping the denominator in factored form. 19. 1 .l. 15*3*m Subtraction of rational numbers can only be done when the fractions have the same denominators. Again we usually get the LCD when the denominators are different as this gives us smaller numbers to work with. sgbtraction of rational numbezs. If a. b and c are rational numbers and b i 0. then a 0 i=2. b b b Perform the indicated operations and simplify. 1 (”3‘5 (NJ-...}. 6 14 (e) 4-1 12 Division of rational numbe s. If a, b, c and d are rational numbers. and b. c and d are not 0. then ab-_-a=m c d b c c 151. (a) l 2 (b) 3 (c) 7/4 (d) 2/21 (9) there is no reciprocal °i=.l-_O,= 4 12 in this case. A reciprocal is a number which when multiplied by the original number equals 1. What number multiplied % - 3a - 151. 152. by 0 equals 1? There is no such number. (Refer to frames 10? - 109 if necessary.) 152°g.é3.=.?g:.6.3.=3. 7 12 7'12 2 153. The denominator of the original fraction was 4/5. ‘We multiplied the numerator by 5/4. These numbers - 4/5 and 5/4 are called the reciprocals of each other. Reciprocal If a and b are rational numbers, then a and b are reciprocals if a'b = 1. Give the reciprocals of each of the following. (a) 2 (b) 2. (0) it 3 7 (d) 21 (e) 0 2 If we look again at the defini- tion of division. we see that dividing by a number is the same as multiplying by the reciprocal of the number. 7 means to multiply 2/3 by the 4 5 reciprocal of 4/5 To multiply two rational numbers. remember we take the product of the numerators of the fractions and divide by the product of the denominators of the fractions. To multiply g by 63 using the 7 12 definition of multiplication. we would have a product of lgé. 84 This fraction is not in its lowest terms. To write an equivalent fraction which is in its lowest terms. we must divide 153. £2.28 = 3.3.3.20202 = 8.81 2020203030303 .Z_._ 2. 2'3 6 (Note - we usually get the prime factors when we are to find the largest common factor.) - 35 - 154. both numerator and denominator by their common factors. It may take several steps to find all the common factors of 126 and 84. Since we obtain equivalent fractions by dividing both numerator and denominator by a common factor we could do this before we multiply. For example. g,- 63,: 6 _ 7 12 7 12) 7'2‘6 Now it is evident that both the numerator and denominator have common factors of 14 and that the equivalent fraction of 9/6 has a common factor of 3 in both numerator and denominator giving a final fraction of 3/2. Multiply 27/8 by 28/81. Perform the indicated operations. Express all fractions in lowest terms. (b) 32.1.22 72 ° 24 (c) ‘ ’ lO igzg (d)1°2l:._1. 3 16 ° 32 (e) éi;;.i2 ’ 19 21 ' 9 27 (f) é&.;.(32.' lg. 21 ' 9 27) 154' (a) 161(285 =54 1000032) 132. or 100 (132) 2%. r% of a number c is a number N such that ....I.‘ (c) = N. (d) 125 .; _l.i = 125(199.) = 100 100 15 In finding 30% of 15. we know the 2 00 values of two of the three 3 quantities in the statement (3)125éld=125;_;5—= 4(c):N. 100 ' 1000 100 25000 ‘We know the values of and 3 f ]_ .... £5 = l .3. .45— : . ( > 25 ‘ 100 25 ‘ 10000 2 0000 3 170. r and o 171. r = 30 and c = 15. when finding 30% of 15. Since we know the values of two of the quantities. we can find the value of the third quantity by performing the indicated Operation. What would you do to find the value of N in the above problem? Find N. 171. Multiply Egg-by‘l32 172. In the formula i6§»(c) = N, we have N a 66 a case where the product of two numbers equals a third number. 16%.and c are called factors. (see frame 17) 172. 173. 174. 175. 176. ..Lpl- Divide the product by the 173. one factor. we need to find the value 174. of c we can divide the value of N by the value of $55.. .lg. = Since 100(0) 24 then 175. =11;_l7.. c 2 ‘100 c=24.:__l2. 176. (a) In figs) = N, c is 177. what we wish to find. 35. = = loo(c) 18, so c 18.1.2.5. = . 100 or c 72 Suppose we knew the value of only one of the factors and the value of the product. how could we find out the value of the other factor? Be specific. If we know that 12% of some number is 24. what uantity in the formula 16%): c) = N do we need to find and how would we find it? Problem: 12% of some number is 24. Set up the division problem to solve for the number. we made use of the principle that if the product of two numbers is a third number. then the third number divided by one of the factors must equal the other factor. Find the value of a number such that 12% of it is 24. (a) Find a number such that 25% of it is 18. (b) Find a number such that 120% of it is 72. (c) Find a number such that it is 22% of 50. (d) 39 is what % of 52? How did we decide in part d of the last frame that r = 75%? We had the statement __r_ _ 3 . 100 - 4 we know that two fractions are equal if the numerators and denominators are equal - in other words. if a = c and b = d and b and d aren't 0 then a,_ b _ 2. d O 177. 178. 179. (b) in -z(c) = N. we wish 12%). 7 100 to find C. 100 1.1-2.0. ' 100 (c) in 105(0) = . .22. = find v. loo(50) N so c = 72 or N = 11. - 42 - and c=60. N. we wish to (d) in 155(c) = N. we wish to r = 9. find r. 150(52) 3 J: ; Thus 100 39 . 52 4:3 2 or 100 4 and r 3.. .25. u 100 75. ..2.-..Z§. - 100 “ 100 and r " r must equal 300 divided by 4, In -—3(c) = N. we wish to 100 find , ..z. = r 100(180) 36 or 43$ 100 180 100 1‘: =0 180 or r 2 SO. 20% of 180 equals 36. 75% 178. 179. 180. ' ..2.-.3 So if in the statement 100 _ 4 , we find a fraction equivalent to 3/4 with a denominator of 100, then the two fractions will have the same denominator and so the numerators must be equal. 3.- .l.. ..EH.._1. = u “ 100 5° 100 ’ 100 and r We also know that two fractions 3 and g are equal if b and d b d aren't 0 and if ad = be. Applying this statement to ..2.=.3 100 4 ’ This states that 4 multiplied by some number r is equal to 300. How would we find the value of r? we get 4(r) = 3(100). What percent of 180 is 36? In frames 167 through 179 we have used the same relationship to solve these problems regardless of which quantities were given. 'We used the relationship i6§(c)=N to find a number N such that it is r% of a number 0. If values of N and r or if values of N and c were given. we also used that if the product of two numbers equals a 180. (a) LCD is 30. 8.1.+z.6.__91 30 was s as (°)3— .3. figsuo (Mg—6° §§=1 ( e) 100 (150) = 9 -43.. number N. then that number divided by one of the factors must equal the other factor. Perfbrm.the indicated operations and simplify. (a) .21 15= .3 .42 (b)38#116 (C>Z.‘£‘ .132 9 '39» (c036- 63 : 1+9 (6) 6% of 150 Chapter 2 - Signed Numbers We stated in frame 86 of the last chapter that the difference of 2 and 3 was written 2 - 3. long as we have only the positive rational numbers and 0. We also stated that this doesn't exist as In this chapter we shall introduce the negative rational numbers and consider the operations of addition, subtraction, multiplication and division with both positive and negative numbers. 1. negative 2. positive 5- a number which can be expressed as the quotient of two integers where the denominator isn't 0 Or 1. 5* O\ o A number preceded by a "-" (negative) sign is called a negative number. - %, - 3, and - of % are examples numbers. A number with a "+" (positive) sign before it is called a W. + 3. + g and + g are examples of numbers. When we talk about negative numbers we must always use the "-" or negative sign. To denote positive numbers, we don't have to use the "+" or positive sign. Thus "2" and "+2" represent the same number. Do "4" and "-4" represent the same number? Consider "4" and "-4". Which represants a positive number? In Chapter 1 we talked about the nonnegative rational numbers. A rational number is In your definition. be sure you stated that the denominator couldn't equal 0. Why can't the denominator equal 0? -h5_ a number of the form a where b a and b are integers and b ¥ 0. because division by 0 is 7. undefined or because you can't divide by 0 or because something divided by 0 doesn't repreSent a number. integers 8 negative integers 10. 0 11. Note: 0 is sometimes referred to as an unsigned number as 0 is neither positive nor negative. parts b and e. In part c. g = 0 12. and - 2% = -6, so % , 15 and ‘ aLL-t'represent integers. Some rational numbers can be expressed in the form a . 1 Rational numbers of this form are called . The positive integers are l. 2. 3. ... the three dots mean and so forth) -1. -2. ~3. ... compose the set of , (two words) Do the sets of positive integers and negative integers include all the integers? What one(s) isn't included in either of these two sets? The set of integers is composed of all integers, i.e., the positive integers, zero and the negative integers make up the set make up the set of integers. Which of the following are sets of integers? (a) 2. 4+. 2% (b) 09 ’37 52 (c) 1 §. .34, 60% (d) ’21 g 9 l. 2 -21: (e) 5 u 157 4 Did you say that part d of the last frame represented a set of integers? 5 isn't an integer. 0 In fact, we can make an even stronger statement about 5 . - M6 - 12. 5 isn't any number. 0 (See frames 10? - 108 in Chapter 1 if necessary) 13. rational l4. rational numbers because they can be expressed as the quotient of two integers where the denominator isn't 0. They are rational numbers and not integers because of 4/3, 715%, and others. 15. - 16. 0 4. (-5) The commutative law states that the order in which two n111nbers are added or multi- Plied is immaterial. (frame 27 Chapter 1) 13. 14. 15. 16. 17. What can we say about 5 ? O The set of integers are part of the set of numbers. .. _ 3. - _ .1. - 6, L} r 3 9 ll! 62 v 15%! 4.3, and O are all examples of numbers. Why? In Chapter 1 - frame 84, we called 0 the identity element of addition because when 0 is added to any number the sum is that number. We could write a + 0 = a. We only stated that this was true for the nonnegative integers. It can be shown that 0 is the identity element of addition for all rational numbers. We shall accept this without proof. Thus, we can say that (-6) + 0 = In Chapter 1, we also stated the commutative, associative and distributive laws for nonnegative integers. We could prove that these laws hold for negative integers and for rational numbers which aren't integers and there- fore for all rational numbers. We will accept these laws for all rational numbers without proof. By the commutative law, (-6)+0= In frame 15. we have (—6) + O = -6 becauSe 0 is the identity element of addition. In frame 16. we have (—6) + 0 = 0 4 (-6) by the commutative law of addition. Thus we may say that 0 + (-6) = - - 47 - Find the following sums: (a) (-17) + 0 = (b)0+24= (0)04-0: (a) 0 + (-21) = 17. (a) -17 18. Sometimes we shall be interested (b) 24 in considering the value of a (c) 0 signed number without considering (d) -21 the Sign of the number. This is called the absolute value of a number. The absolute value of -2 is 2. The absolute value of +2 is 2. What is the absolute value of —6? 18. 6 19. Find the absolute value of: _ l (a) a (b) ' 003 (e) + Z 3 (d) - 425 (e) 42 19. (a) g 20. What is the absolute value of +% ? (b) .03 What other number has the same absolute value as +2 ? (e) Z 3 (d) 425 (e) 42 20. absolute value of +l is l . 21. What number has the same absolute l 2 2 value as -7? - E'has the same absolute Ealue as 4 % . 21° *7 22. The absolute value of a number b is written b . -4 is read . 22. the absolute value of -4. 23. I3 - 2] is read . 23. the absolute value of thg 24. To find the value of '3 - 2'. we number (3 - 2). must first find the value of 3 - 2 and then find the absolute value of this. l3-2l=l 1= 24. I3 - 2' = Ill = 1 25. Does l3] - [2| tell us to follow the same procedure as 3 - 2 25. no 26. What roc dure are we asked to do in|3 - 2|? 26. find the absolute value of 27. Express the following without 3. then find the absolute absolute value signs and simplify. value of 2 and then find the difference of the results. (a) l+4| + I-4I (b) M! — I-ul 1 1 - .. + _ .. (e) l 2| I “I l l d _ - - (>12 “I (e) 12 + 3(4)] 27. (a) 4 + 4 = 8 28. Two numbers have a sum of 0 when (b) 4 - 4 = 0 one number is a positive number (c) I + I _ 3 and the other number is a negative 2 4 — 4 number and both have the same (d) ll _ ; absolute value. 4 ' 4 (e) 2 + 12' = I14! = 14 For example. the sum of +4 and -4 is 0 because +4 is a positive number and -4 is a negative number and both numbers have the same absolute value. The sum of -31 and 31 is 28: 0 29. The sum of -31 and 31 can be written as -31 + 31 where the 4 is used to mean add and is not the sign of the number. What number added to 16 gives a sum of 0? In other words. 16 + = O. 29. -16 1. .1 30. (a) 2 or + 2 (b) 14 or +14 (c) -35 (d) 0 (e) 0 31. :21- or + i- because the sum of - .1. and 4- .1. is 0 2 2 32. -29 because the sum of 29 and -29 is O 330 '3 negative 34. their sum.must be 0 35. 14 or +14 because the sum 0f -l4 and 14 is 0 -49.. 30. 310 32. 33. 34. 35. 36. Complete: (MO-320+ =0 (b) __ + (-14) = 0 (c) (d) ~10 + 10 = (e) 16 + (-16) = __ The numbers 16 and -16 are called the additive inverses of each other. Definition: One number is called the additive inverse of another if their sum is O. +35=o What is the additive inverse of - %.? Why? What is the additive inverse of 29? Why? Numbers which are additive inverses of each other are also called the negatives of each other. The additive inverse of 3 is and so -3 is the of 3. What must be true of a number and its negative? The negative of -14 is because . What is the negative (the additive inverse) of each of the following? (a) a. (e) ~1§ (b) 34 (r) 0 ~23 (s) (5+2) (d) - t- (h) - <4-3) 36. 37. 39. 40. (a) -1% (b) -34 (e) 23 (d) .1; (e) ‘5'- (f) 0 (g) - (5+2) (h) + (4-3) +(-2+4) or (-2+4) because -(-2+4) + (-2+4) = - -1, ..a, - 23. 4 3 (1.3) additive inverse (see frame 31) yes - 5o - 37. 38. 39. 40. 41. In part f of the preceding frame 0 is the correct answer. “we do 39;,use a "-" sign as 0 is neither positive nor negative. In parts g and h we have only one number given. This is shown by the use of parentheses. Parentheses are used to show that several terms are to be considered as one number. (5+2) is the same as + (5+2) or is a positive number and the negative of this number would be - (5+2). What is the negative of -(-2+4)? Note that we have used the word negative in two different ways. First. we talked of a negative number. In frame 36 which of the given numbers were negative numbers? Negative numbers are numbers preceded by a "-" (negative) sign. Secondly. we talked of the negative of a number. Another name for the negative of a number is the O In frame 36 part a, we found the negative of \ was - 2.. Here the given number2 is a positive number and its negative is a negative number. Will this be true for the negatives of all positive numbers? In frame 36. we found that the negative of -23 was 23. In this case we were given a negative number and the negative of this number was a positive number. What can you say about the negative of a negative number? - 51 - 41. it will always be a positive 42. number ”2. 35 - ~35 or 35 = - (-35) 43. 43. not necessarily -b is a negative number if b is positive and -b is a positive number if b is a negative number Mo--5=5 1+5. _ = i 45. (a) - - (b) 1 (e) 03 In symbols. the statement "23 is the negative of -23" would be written: 23 = - -23 or 23 = - (-23). Write in symbols: negative of —35. 35 is the If h represents some number. then -b represents the negative of this number. Does -b repreSent a negative number? Explain. Now we can give the definition of absolute value. The abs lute value of a number b (written lbl) is: lb] = b if b is a positive number or is 0 Ibl -b if b is a negative number Thus. [-3] - -3 as the given number -3 is a negative number. l—5l = In other words, the absolute value of a positive number or zero is the number itself and the absolute value of a negative number is the negative of the number. Complete: (a) l- gl = (b) 113! = (c) ‘0' = (a) If a represents a positive number. then a and this is a (positive. negative) number choose one (b) If e represents a negative number. then c and this is a (positive, negative) number oose one - 52 - 46. (a) a positive (b) -c positive 47. absolute value of -2 2 3 48. -5 #9. (a) - g. (M83 5 47. 49. 50. 50. find the absolute values of 51- the numbers and then take the feronce of the absolute values. ‘We will accept without proof that the commutative. associative and distributive laws hold for all rational numbers. Therefore. we may make use of these laws in doing computations involving both positive and negative rational numbers. Addition 9f Signed Nggbegs a To add two numbers having the same sign. add the absolute values and use the common sign. Tb add two numbers having different signs. take the difference of the absolute values and use the sign of the larger number. (b) For example. add -2 .:3_. 'We will use part a above since these numbers have the same sign. "2' meals 0 . i-3l = _____.. :.(&.means therefore) adding the absolute values and using the common sign. the sum of -2 and -3 is . l—2l = (a) Find the sum of - % and - %'° (b) Add: 2 3. 65 Tb find the sum of -3 and +2. we need to use part b of frame 47 since the numbers have different signs. 'We first have to .1. . The sum.of -3 and 2 is __. . we used the "-" sign because - 53 - DON'T say subtract them as you will later confuse the operations of addition and subtraction. 51. -l 52. because we use the sign of the number with the larger absolute value 52. sum is h or +4 53. found the absolute values. then found the difference of these. and used the sign of the 20 as it has the larger absolute value 53- -27 51‘. - 3 - 7 -34 1 54. 8 5 55. 55. (a) + 56, (b) + (c) -11 (d) -6 (e) 12 or +12 gr; '4: g ' 5 56. negatives 57. or additive inverses Add: ~16 ___ZQ__ Explain how you found the sum. Add: 16 o .44 -21 :&3 :1 __Z 2;: The sum of -6 % and 15 is Complete the following. (a) -13 _______,9 = -4 (b)2_____-ll=—9 (c) 5 + = -6 (d) + -9 = -15 (e) + -23 = -11 (f) u + _________. = 0 3 = o (g) _.______ + 5 In 4 + -4 = 0. 4 and -4 are known as the of each other. (See frame 33 if necessary) % is the negative of *. 57. - %_ 58. What must always be true of a number and its negative? -54- 58. The sum of a number and 59. 590 60. 61. 62. 63. its negative is always 0. -3 60. -15 Both.of them tell us to 61. add. addition 62. ‘12 630 8 (Remanber positive numbers can be written without the + sign) associative 64. If we have more than two numbers to add. we can make use of the associative law. For example. Add: -12 _-:.L9_ we can add -10 and 7. getting a sum of . and then add this to ~12. getting . The example. Add: -12 7 ..zl9__ can also be written as: -12 + 7 + -10. Do the + signs here tell us to add or do they tell us that we have positive numbers? Note that if there is only one sign between two numbers. it tells the operation to perform. If there are two signs between two numbers. the first tells the Operation and the second is the sign of the number. In 8 + -12. what Operation is to be perfbrmed? In 8 4 ~12. is a negative number and is a positive number. Add: ~12 7 __:lQ_. ‘we said you could have added the ~10 and 7 and then added the result to -120 This makes use of the law. ‘we also could have added -12 and 7 and then added the result to -10. 64. 65 66. 67. w. 69. (-22 +50) 4- -28 = 28 + -28 = 0 or -22 4- (50 + -28) = -22 + 22 the absolute value of -5 (see frame 44 if necessary) 5 + 9 + 2 = 16 l-2|+-7=2+-7=-5 find the absolute values of each number - 55 - 65. = O 66. 68. 69. 70. 70- l-Zl + l13l + l-21l + l-5l = 71. 2+13+21+5=41 The associative law states that (-12 + 7) + -10 = -12 + (7 + -lO). or in other words that the grouping of terms in addition is immaterial. Use the associative law to complete: -22 + 50 4 -28 = Complete the following. (a) - 3/4 + -9/4 + 5/4 = (b) 1n + -3 + -2 + 12 + -8 + 16 + -7+15= (c)12+-l5+-13+2o+-55+ -15+4+6= I-SI means . l-5l + l-9! + 2 = = l-3 +1! + -7 = ...... = Emample: find the sum of the absolute values of -2. 13. -21. -5. In this problem. you are asked to do two things. One is to find the sum or to add and the other is to find the absolute values. In a case such as this. you have to decide which you are told to do firSto Which do you do first in this case? Find the sum of the absolute values of -2. 13. -21 and -5. If you are asked to find the absolute value of the sum of -2. 13. -21 and -5. are you asked to do the same process as in the above problem? -56- 71. no 72. Explain the procedure you are asked to use. 72. Add the given numbers and 73. Find the absolute value of the sum then find the absolute value of -2. 13. -21 and -5. of the result. 73. L2 4. 13 4- -21 + -5] :2 7’4. You must always read a problan carefully to see exactly what you l-ljl = 74. are supposed to do. Sometimes it will make a difference in the 15 answer you get. (a) Find the absolute value of the (b) Find the sum of the absolute values of -32 and 21. 74. (a) [-32 + 21' = l-lll = 11 75. In frame 86 of Chapter 1. we defined subtraction for nonnegative (b) l-32l + I21! = 32 + 21 = 53 integers. We can define subtrac- tion of all rational numbers in the same way. Sgbtraction If a and b are two rational numbers. then a-b represents their difference and a-b = c only if c¢b = a. By this definition. 3 - 4 = --1 because --1 + 4 == 3. -6 - = 2 because -6 75. -8 belongs in both places 76. Complete the following. 24- (a) -6 - __ = -10 because (b) 0 - -3 = __ because (c) -3 - = -3 because 76. 77. 78. 79. 80. 81. 82. 83. 84, -57- (a) 4 or +4 because -10 + 4 = 77. 6 (b) 3 or +3 because 3+-3=o (c) 0 because -3 + 0 = -3 +11 or 11 subtract It tells us that we have a negative number -5+-7 aid -5 and -7 We get -12 l3‘+ +# = 17 subtract 4 or +4 from -3 78. 79. 80. 81. 82. 83. 84. 85. Doing subtraction by using the definition is not a very rapid process. It can be proven that a-b=a+(-b)orthat subtracting two numbers equals the sum of the first number and the negative of the second. Then-5--ll=-5+ We have two signs preceding the 11. The first one tells us to What does the second sign mean? -5 - -11 tells us to subtract -11 from -5. According to frame 77: this can be done by adding the negative of -11 to -5. or -5 - -11 = -5 + +11 Note that we found the negative of ~11 and also changed to the operation of addition. -5-4-7: To complete the problem -5 - +7 = -5 + -7. we must now (what Operation on what numbers?) getting 0 u-_u= In -3 - u. there is only one sign between the numbers. This tells us to . In -3 - 4. we are to subtract from . “We are ngtlto subtract a -h. If you said -4. you have considered ~3--‘+19.t-3-4. 85. 86. 87. 88. - 58 - -3-43-34—142-7 86. -16 *,,5 = ~11 87. Remember you are adding two numbers which have different signs so you find the difference of their absolute values and use the sign of the number with the largest absolute value. (See frame 47 if you need to review the addition of signed numbers.) addition 88. add -14 and -6 subtract 89. Subtract 11 from -8 or subtract +11 from -8 There is only one "-" sign between the two numbers and if you say that it means to subtract. you have used it and there is no other sign written between those numbers. Making use of the fact that the difference of two numbers equals the sum of the first number and the negative of the second number or that a-b = a + (-b). we can write-3-h=3+ = Change to an equivalent addition problem and simplify. -16--5= .-. In -lh + -6. what operation is to be performed and on what numbers are you to operate? In -8 - 11. what Operation is to be performed and on what numbers are you to Operate? Evaluate the following. (a) -7 - -# = w>e-3= (c) O - -2 (d)-u - -9 = (e) 13 - -21 = (f) -18 + -12 = (g) -26 + 17 = (h) 35 - -6 = (i) #2 - 59 = (j) 34 + -16 = 89. 90. 91. 92. 93. (a) -7 + +4 = -3 (b) -8 + -3 = ~11 (c) 0 + +2 = 2 (d) -4 + +9 = 5 (e) 13 + +21 = 3n (f) ~30 Careful here. Did you notice that this was addition? _ 59 - 90. 91. (g) -9 m)%++6=fl (i) 42 + ~59 = ~17 (j) 18 -7 + -5 ~ -u = ~7 + -5 + ii -7 + -5 - -4 = ~7 + ~5 + +4 = 92. (-7 + -S) + u = ~12 + 4 = -8 or ~7 + ~5 + +h -7 #'-1 = -7 +(-5 + +4) (a) ~19 + +6 = ~13 (b) ~11 + ~15 = ~26 (0) -16 + -5 + 42 = (~16 + ~5)+2 = -21 + 2 = ~19 (d) (#3 + ~32) + 457 = 11 + 57 = 68 (6) Remember parentheses are used to indicate that a group of terms are to be used as a single number. 93. so first we must evaluate {-17 + 8). Then we must evaluate (5 - 15) and then subtract these numbers. (-17 + 8) - (5 - 1n) = -9 - -9 -9 + +9 = 0 (a) -(-9 ~ 6) + (7 - ~8) = - ~15 + 15 = 30 (b) - (12 ~ 25) + (n - 11) ~ (~21 + 32) = ~ ~13 + -7 - 11 = + +13 + ~7 + ~11 = 94. -5 If you are to combine ~7 + -5 - -4. you can use the associative law if the ogly;operation to be used is addition. In ~7 + -5 ~ ~h you are to add the first two numbers but then you are to subtract ~4. ‘Write ~7 + ~5 ~ ~4 so that it involves only the operation of addition. -7 + -5 - ~4 = ~7 + -5 + Now use the associative law to find a single number which expresses the value of '7 + "'5 "' '40 Complete the following. (a) -19 - -6 = (b) ~11 ~ 15 = (c) ~16 ~ 5 ~ -2 = (d) 43 + -32 - -57 = (e) (~17 + 8) - (5 - lh) = Find the following values. (a) - (-9 - 6) + (7 - ~8) = (b) ~ (12 ~ 25) + (4 ~ 11) ~ (~21 + 32) = (0) ~41 ~ (11 ~ 32) + (~1u ~ ~20) Multiplication of signed numbers To multiply two signed numbers. multiply the absolute values. The product is positive if the numbers have the same sign. The 94. 95. 96. 97. 98. (c) ~41 ~ (11 ~ 32) + (~14 ~ ~20) = -41 - ~21 + 6 = 26 .6 This time the product is a negative number since the two numbers have different signs. 30, -66. 4+. -63, “'2' 0’ 0 because any number multi- plied by 0 equals 0 one of the numbers or else both.of them must equal zero Dondt forget the case where both.numbers equal 0. either a or b or both must equal 0 - 60 - 96. 97. 98. 99. product is negative if the numbers have different signs. If we are to multiply ~6 and ~3. we would first find the absolute values of these numbers. The absolute values are 6 and 3 respectively. The product of the absolute values is 18. Since the two numbers have the same sign. the product is a positive number. Hence. the product of ~6 and ~3 is 18. What is the product of 2 and ~3? Multiply: ‘1 -5 ~11 2 9 -7 0 -4 .412 ..é. :§. :2. '6 :3. .9. Why were the products in the last two problems equal to 0? If the product of two numbers is zero. what can you say about the numbers? we can state: if a and b are rational numbers such that ab -.-.- 0. then at least one of the factors is 0. What does it mean to say "at least one of the factors is 0"? Remember we agreed not to use x to mean.multip1ication. Instead we will use either a dot or parentheses. It will be much less confusing if we use parentheses when we are dealing with negative numbers and mean multiplication. (a) ~5<~7> = (b) 9 (___> = ~63 (e) -2(____) = -2 (a) <- ex-..) = 100. 101. 102. 103. lOfi. 105. (d) 7 or +7 factor 5 and 7 ~1 and ~35 1 and 35 These can be listed in any'order. ~3 and 2 3 and ~2 ~l and 6 1 and ~6 2(2)(2) or three factors of 2 (~2)(-2)(-2) -8 -61.. 100. 101. 102. 103. 104. 105. 106. In (~5)(~7) the ~5 is called a Since (~5)(~7) = 35. we could say that the factors of 35 are ~5 and -7. List three other different sets of factors of 35. What are 4 different sets of factors of ~6? 23 is a shortcut way of writing (~2)3 is a shortcut way of writing . --23 does £19; mean the same as (-2)30 ~23 is the same as (~1)(23). YOu will note that 0311 the two is raised to the third power. 'Written in terms of factors. ~23: (~l)(2)(2)(2). To evaluate this. we make use of the associative law and find that -23 = (-l)(2)(2)(2) = ‘We said that (~2)3 = (-2)(~2)(~2) and this also can be evaluated using the associative law. (~2)3 = (~2)(-2)(-2) = 106. -8 10?. (-3)(-3) = 9 108. (~1)(3)(3) = -9 Note that the results 0 {-3)2 and -32 are 221 equal. 109. (~1)(2)(2)(2)(2) = ~16 111 (a) (-5)(-5)(-5) = -125 (b) (-1)(3)(3)(3)(3) = (e) (-3)(-3)(-3)(-3) = (d) (-l) (4)(4) = -16 (e) (-4)(.u) = 16 1(1)(l)(l)(1) = l or five factors of l which equals 1 f - 62 - 107. 108. 109. 110. 112. Some of you will wonder why the emphasis on the meaning of the two quantities. after all the answer came out the same. It did in this case. but this doesn't always happen. Even if it does. the important thing is to do exactly what the symbols tell you to do. (~3)2 means and this equals . -32 means and this equals . Look carefully at (~3)2 and ~32. (~3)2 tells us to square the number ~3. -32 tells us to square the number 3 and then to multiply the result by ~l. -24 = _ Evaluate. (a) <-5>3 (b) a“ (ex-3)“ (d) -42 (6) (4+)2 15 means _____________.and this equals (-l)5 means and this equals . 112. (-l)(-l)(-1)(-l)(-l) = 1 113. 114. 115. 116. 117. 118 - 63 - 113. 114. -8) 9) = -72 ~2) = ~32 -27)(8) = ~216 ~1)(25)(16) = .400 -9)2 = 81 115. ~2~3 means to subtract +3 from ~2. ~2(~3) means to multiply ~2 and ~3. multiply 3 by'~4 and then 116. subtract the result from 2. yes because subtracting 117. two numbers is equal to the sum of the first and the negative of the second. In 2 ~ 3(~4). the second number is 3(~4) and its negative is (~3)(~h). Hence. 2 -3(-4) = 2 f (~3)(-4). 16 + ~18 + 1 = ~1 (~9)(5) = -45 118. 119. Simplify the following: (a) 3<4><9><~1> = (b)(-2)3(3)3 = (c) (-9+7)5 = (d) (-3>3(2)3 -- (e) _52(_2)4 = W>GB+4F= Do ~2~3 and ~2(~3) mean the same? Explain. Tb simplify 2 ~3(~4). what procedure would you follow? -3(-4) 2 ~ ~12 (Multiplying 3 by ~4) 2 + +12 = 1# (Performing the subtraction) Could 2 ~ 3(~#) be considered as 2 + (~3)(~4)? Explain. Then in evaluating 2 ~ 3(~4). we could have multiplied ~3 and -# and then added the results. Simplify: “2 + 9(~2) +1 {-3 + -6)(23 + -3) (l3 4 ~6!)3 = Simplify: - 64 - 119. (l-3|)3 = (3)3 = 27 120. 120. [-56 + 62][19 - +33] -- (6)(~1h) = ~84 121. 121. - f}? (~lu)(~8[-2] + ~6) = 122. ~%(~14)(16 + ~6) = 7(10) = 70 122. because division by O is not defined or because we can't divide by O 123. ~9 because (~9)(2) = -18 124. 3 because 3(~13) = -39 9 because 9(~6) = ~54 ~9 because ~9(~8) = 72 1230 121+. 125. [(-8)(7) + 623[19 ~ (-11)(-3)] = - 9g- (-2 - 12)<-8[15 ~ 17]. -6) = In Chapter 1 frame 96. we defined division of nonnegative integers in terms of multiplica- tion. We can use the same definition for all rational numbers. Quit-Ra If a and b are rational numbers and b i 0. the quotient of a and b is written g_and §.= c b b only if c'b = a. Why can't b equal zero? Then to divide ~18 by 2. we must find a number such that when it is multiplied by 2 we get ~18. :1§.= 2 because ~39 divided by ~13 = because . ~54 divided by = ~6 because . 72 divided by = ~8 because . By the examples in the last frame. you can see that the division of signed numbers follows the same rule as that for multiplication of signed numbers. Division of signed numbers To divide two signed numbers. divide the absolute values. The result is positive if the numbers have the same sign. The result is negative if the numbers have different signs. Find the following quotients. :6_§.-6.1.35._9.:£2 ~17 1 ~9 ~2 8 - 65 - .7. mg 11 ~21 -52 O 125. 4 126. Are "find the quotient of ~68 and -6 -17" and "find the quotient of -17 ~15 ~17 and ~68" equivalent state- 0 (Every fraction with a ments? Explain. zero numerator and a number which is not zero in the denominator equals 0) 5E” z. -; 3 - 39 13 no number (Remember division by 0 is net defined) 126. no. the first means 2é§ 127. Since the order of two numbers is “17 important in division. we may say the second means :ég that division is not ____________. 127. commutative 128. (a) and (To show falseness. one of rational numbers are counter example is commutative operations. sufficient). (b) and of rational numbers are not commutative operations. 128. (a) multiplication and 129. Perfonn the indicated operations ad on and simplify. (b) subtraction and division (a) - + 10 (b) 12 ~ ~16 11 ~ 18 (c) 513 ~ 20?(12 + 2) ~8 - 3 1 ~ 51 129. (a) :g; = -7 130. If we are given any two rational numbers. then one of three (b) 3; = -5 conditions exists. Either the two ‘ numbers are equal or the first (c) g-z)§22) = _ g number is larger than the second i -11 -35 5 one or the first number is smaller than the second one. 131. -u - -2 -2 - -4 -2 - 66 - 131. 132. The signs of inequality are > and <. If a and b represent two rational numbers such that a is larger than b, we may write a>b. Of course, if a is larger than b, then b is smaller than a and this is written ba, if b - a is positive. To decide whether -4 is larger or Smaller than ~2. we shall evaluate the difference of -h and -2 and we shall also evaluate the difference of -2 and ~b. -4 - -2 = -2--u= The difference of -2 and -4 is positive. which according to our definition means that -2 is the largzr of the two numbers. or that -2>- . Which is the larger: -1 or -5? write a statement showing this relationship. - 67 _ 132. -l is the larger because 133. Complete the following using =, -l - -5 is a positive > or <. number. (a) 2 h l. -l>-5 or -5<-l (b) 2“ 2 z. (c) g- 5. (d - 13. - ll. ) 3 3 133. (a) < 131+. Suppose that we were to determine (b) = the larger of 5/6 and 7/9. (c) > Using the definition. we would d < evaluate g- - g and g - 2. to find out which difference was positive. In order to find out the value of g- - g- what procedure would you have to follow? 13“. In order to add or subtract 135. Simplify'2-- 2., fractions you need the same 9 denominator so each fraction must be changed to an equivalent fraction having the same denominator. 135. LCD is 18. 136. Since the difference is a positive _ Z = l: _ 1.3 = _2L number. is 2 or Z the larger? 9 18 18 18 9 136. 5/6 137. Note that if both fractions are positive numbers. the larger fraction has the larger numerator when the denominators are the same. Using =. > or < complete the following. 3. 39. (a) 5 65 (b).;3 lg 22 77 .. _Z - 11 (c) 60 8a -68- l-2l l-5I 2 6 (d) l-3l (e) la! m l- 3| I-a —% =>>><=> Chapter 3 - Operations with Polynomials In the last two chapters we studied the rational numbers and the operations of addition. subtraction. multiplication and division on these numbers. In this chapter the Operations of addition. subtraction. multiplication and division on polynomials are studied. In this chapter and in fact in the next three chapters. all the letters that we use will be understood to denote rational numbers. 2. 15(- 25'?) or -3 3. 10. 20, 30, no 4. variable 5. 31.513 represents 150 if p is 10 5p 15p 15p 300 if p is 20 ’+50 if p is 30 600 if p is 1&0 1. We will use letters to stand for 6. rational numbers. If you are given that p is a rational number. does p represent any specific rational number? 15p is a symbol which we will use to mean 15 multiplied by the value of the number p. If p stands for the number 3. then 15p represents the number 15(3) or #5. If p represents the number - l/So then 15p represents the number 01‘ o In 15p. the p is called a 22212212. A variable is a symbol which represents one or more numbers in a given set of numbers. If p represents any integer less than 50 which is divisible by 10. list the possible values of p. possible values of p are Since p represents one or more numbers in a given set. p is called a _ If p represents any of the integers 10. 20. 30 or 40. what does 15p represent? b2 is read "b squared" and means (b)(b) or b'b. b3 is read "b cubed" and means -69- 9. 10. 12. 13. 14. b3 means b-b~b. three factors of b b4 means b°b°b°b x5 x to the fifth power 24 3(3) exponent of the base b. a(a)(a)(a)(a)(b>(b)(b) there are 5 factors of a and 3 facunnsof'b one five two one three - 7o - 7. 9. 10. ll. 12. 13. 14. 15. or that there are _________factors of b. hOW'many bu is read "b to the fourth power" and means . (x)(x)(x)(x)(x) can be written which is read (2)(2)(2)(2) can be written 32 means . In x6. the 6 is called an egponent and x is called the base. In a5b3, the 3 is called the of the base . a5b3 means . uxjyzmeans that there is _____ factor of 4. factors of x and factors of y; pq3 means that there is factor of’p and factors of q. (ab2)# means that there are four factors of . Note that in the above. the exponent applies to the letter. symbol or number which.immediately precedes it and to no other letter. number or symbol. H r 15. 16. 17. 18. 19. 20. L - 71 - n factors of a 16. or a'a'a°...°a 1?. monomial yes 18. Check the definition. Is this a product of a rational number and variables raised to positive integer powers? Remember () are used to indicate that a group of terms is to be used as a single number. Thus one variable is (xsy) and this has an exponent of l and the variable (a+b) has an exponent of 2. No. y2 is divided by x and 19. not.multiplied by it. Lg'XA 9 ~2mn5. and -abc are 20. monomials. 3P-2 doesn't have a positive integer exponent. “Bxfl8p is not a product 1 2' 21. If n is a positive integer. then an is defined to be . 15p. --llm3n2 and 55-572 are examples of monomials. A.monomial is the product of a rational number and a variable (or variables) raised to a positive integral power (or powers). -8r356 is a . Is 5(x:-;y)(a+b)2 a monomial? Give a reason for your answer. Is §2 a monomial? Give a reason for your answer. Choose the monomials from the following. ;Q_ 4 3p'2 ~2mn5 -3xw8p 3X' 9 a o -abc In -2mn5. the -2 is called a ggefficient. A coefficient is the numerical factor in a monomial. In - %a3. the coefficient is In - %a3. the 3 is called an 21. 22. 23. 24. 25. 26. exponent or power coefficient - -.l. .. 2960511) 4’ l commutative associative distributive commutative distributive - 72 - 22. 230 24. 25. 26. 27. In -11n2nt3, -11 is called the In y“ there is no coefficient written however this coefficient is understood to be 1. Thus y“ and 1y“ are regarded as equivalent. What is the coefficient of -a2b? Name the coefficient of each of the following. -2ab, 6x2. 5(c-d). xzy. - inZZprZ, ~w5 Since we are using letters to represent numbers. then addition and multiplication of monomials must satisfy the same rules as addition and multiplication of rational numbers. These laws are the o o 2i and a. laws. The law states that the order of two numbers in addition or multiplication is immaterial. The law connects the two operations of addition and multiplication. The addition of two numbers such as 4x and 7:: may be done by applying the distributive law. 4x + 7x = x(4+7) = x(ll) = 11x 'We generally write the numerical symbol first. Using the distributive law add 3p and 9p. _ 73 - 2?. 3p+9p=p(3+9)= p(12)= 12p 28- 28. by the commutative law of multiplication 29. w2(-l+ + 9 4- -2) = w2(3) = 30 4y+y= y(4+l)= y(5)= Don't forget y is the samey as 1y. 0 31. 6rzs+8rzs+ -2r25 = rzs(6+8+ -2) = 12r25 32' 3Xy2+7xy2+2x2y+x2y+hx2y = m2(3+7)+x2y(2+1+4) = 10W2+7x2y 29. 3W2 3o. 31. 33. Why does p(12) equal 12p? Use the distributive law to find the following sum. --l+w2 + 9w2 + —2w2 = Find the sum of 4y and y using the distributive law. Find the sum of 6r25. 8rzs and Example: Add 2a, 3b, ha, 8a and 5b 2a+3b+l+a+8a+5b= You will notice in this case that there is no number distributed over the whole sum. It is therefore necessary to rearrange the terms and group them so that in each group there is a number distributed over the whole group. We can do this as addition of rational numbers is commutative and associative. 2a+3b+4a+8a+5b = 2a+ha+8a+3b+5b Now a is distributed over the first three terms and b is distributed over the last two tenns. Applying the distributive law to the first three terms and then to the last two terms. we get 2a+4a+8a+3b+5b= a(2+4+8)+b(3+5) and this equals lha+ b, Add 3xy2 . 2x2y. 3623'. 7xy2 and hxzy. 3xy2 and 7xy2 are called similar monomials. Monomials which have exactly the same factors except for coefficients are called similar monomials. Are 3p2 and -7p2 similar monomials? Why? 33. No. p2 and p3 are not the same factor. 31+. 7mn2+ -llnm2+ ~2m2n+6m2n+ -m2n+3m2n2 = mn2(7+ -ll)+ m2n(-2+6+ -l)+3m2n2 = 4+ng + 3m2n + 3:112:12 35. addition 36. polynomial 37' variable is p ghest power is 3 degree of polynomial is 3 -74- 3“. 35. 37. 38. To add similar’monomials. add the coefficients and multiply the result by the common factor. Thus to add -8mn and lZmn. we can add the coefficients of -8 and 12. getting 4. and multiply this result by the common factor of mn, getting hmn. This is exactly what we do when we use the distributive law. -8mn+lZmn = mn(-8+12) = amn. Add 7ng. -2m2n, 6m2n. 3mZn2, --llmn2 and -m2n. -umn2+3m2n+3m2n2 is a polynomial in two variables. A polynomial consists of one or more monomials used as terms. Terms are connected by the operation of . 3m4+5m2-6m is a in one variable. 3p2-4p3 is also a polynomial in one variable. This may not seem to fit the definition as the terms are not added but remember 3132-4103 == 3p2+ 4103. The degree of a polynomial is the largest power of a particular variable. In 3p2-Qp3. the variable is . and the largest exponent of this variable is . so 3p2-l45p3 is a polynomial of degree What is the degree of -6y5+13y3-8y9-y5 7 38. 9 39. 0 because there is no variable present 40. the second term is hx3 its coefficient is h or +u 1+1. 2x7-7xn+5x2+3x+6 “2. multiplication “3. the term with no variable is written first. then the term.with the variable to the first power. then the term.with the variable Squared. etc. - 75 _ 39. 40. 41. #2. 43. 44. We will agree that if a term has no variable, it is of degree zero. If we consider the term 6, it is of degree because . Given the polynomial in x. 3x5+4x3-3x2+7. what is the coefficient of the second term? The polynomial 3x5+4x3-3x2+7 is arranged in descending ordeg. This means that the term with the largest exponent is written first. then the term with the next largest exponent, etc. until finally we have the term without the variable (i.e. the term of degree 0). Arrange in descending order: 3x~7x4+5x2+6+2x7 Be sure you have a 4 or a - between each term. If you wrote 5x? 3x, this means to perform the opera- tion of . A polynomial may be arranged in ascending grder. How would a polynomial be written if it is in ascending order? In the polynomial 3x§+4x3-3x2+7. the terms containing x9 and x are missing. It is often convenient to to consider the polynomial with these terms present. Since any number multiplied by 0 equals 0. we can write 3x5+nx3-3x2+7 as 3x5+0'x9+4x3 ~3x2+0°xw7 In other words, we consider the coefficients of the missing terms to be . - 76 - 1&4. 0 45. Considering the coefficients of these terms to be zero. doesn't change the polynomial because 1+5. 0 multiplied by any number 46. We can rewrite any polynomial or is 0. rational number in a different form providing the value remains Ranember xL" and x represent the same. numbers In 2p3+5p+7p5-3. the coefficient of the fifth degree term is . The coefficient of the fourth power term is . 1&6. 7 or +7 47. To add polynomials. we make use of the fourth power term is missing. so the coefficient is 0 47- <-5p6+7p2-4p3>+(3p6+2p‘*-7p2)= (61064-3106 >+<7p2+-7p2>+(-4p3> +(2p49 by the commutative and associative laws. ’48. By the distributive law. we get. (-5+3>p6+(7+ -7)p2+(-hp3)+(2p“> which equals ~2P6-4PB‘F'2P4 48. (a) ~2a+8b-4e-3d (b) ~8x3+2x-2+9X2-5Xu ’49 . the commutative. associative and distributive laws. For example: Add 31y2+y3-7 and -1+y3+6y2-2y+3. This is: (3y2+y3-7)+(-#y3+6y2-2y+3) Using both the associative and commutative laws. this equals: +<-2y>+(-7+3). Using the distributive law. we get: (14)y3+(}r6>y2+<-2y)+(-7+3) which equals -3y3+9y2-2y-u. Add: -5p6+7p2-l+p3 and 3p6+2pu-7p2. Find the sum of: (a) 3c-4b+7c+d and -5a+12b-11c-4d (b) -x3+4x.8 and 9x2-5x946-7x3-2x To the sum of an-l-xz-x and -5x2-7x3+3 add the sum of 49. 50. 51. 52. 53 - 77 - The terms may be arranged in any order. Why may we say that -2a+8b-#c-3d equals 8b-2a-3d-kc sum of 2xé+x2~x and -5x2-7x3+3 is 2x4—ux2-7x3-x+3. sum of —x3-8 and 4x+9x2—5x9+6 is -x3-2+4xm9x2-5x9. -x3-8 and 4x+9x2-5x§+6. 50. What is the coefficient of -(-2xsy+5p)? (Zxa-4x2-7x3-x+3)+(-x3-2+4x#9x2-5xu) = —3x9+5x?-8x3+3x+1 -1 51. Remember that () are used to indicate that a group of tenns is to be used as a single number. -l(—2x#y45p) = Zth-Sp 52. 5a-3b+4c 53- +(5a-3b+#c) means to mmltiply by l or +1 and any number multiplied by 1 equals itself. 3a54b-2a+8b because -(2a-8b) is -l(2a-8b) and by the distributive law this becomes ~1°23+(-1)(-8b) = 54. -Za+8b o -(3P-L*q) check this by multiplying -l by 3p-hq to see if you get -3p+#q. 55. -(-2x§y+5p) can be written without parentheses by using the distributive law. By the distributive law. -(-2X#y*5p) = Write +(5a-3b+4c) without parentheses. +(5a-3b+4c) = 3a+1+b-( 28.-8b) = when written without parentheses. Insert a quantity inside parentheses which are preceded by a - sign so that the entire quantity equals -3p+4q. -( ) = 6qu If we are to subtract 3x-2 from -2xp4, we would use parentheses around these two quantities to indicate that the whole quantity was to be treated as one number. We would write: (-2x-4)—(3x-2). Without the parentheses. this quantity equals . ~78- 55. -2xsh-3x#2 56. If we combined similar terms in because -2xph-3x»2. we would obtain (-2xeu) = +1(-2xsu) = -2x~4 and . . -<3x-2> = hex-u) = -3x+4 56. -SXh2 57. Write: subtract hmn-sz from -m*3n20 57. (-3mn+3m2) - (“mn-sz) 58. Remove the parentheses and complete the subtraction. 58. ~3mnr3m2-4mn+5m2 = -7mn+8m2 59. write: the difference between xz+fixs2 and ~3x2+6xé7 59. (x?+4ms2) - (3x2+6xw7) 60. Complete the subtraction in the preceding frame being sure to Yen must have the second set combine terms. of parentheses as you are to subtract the whole quantity of -3x?+6xw7. 60. x2+4m+2+3x3-6xe7 = uxZ—zxsg 61. Find the sum of -lla+8b-2c and a-ljb-19c and then find the difference between this result and 9a+b+c. 61. sum of the first two 62. What number added to -3p3+8q2 quantities is -lOa-5b-21c. equals 0? The subtraction is written (~lOa-5b-Zlc)-(9a¥b+c). This equals -lOa-5b-21c-9a-b-c or -l9a-6b-220. 62. You are asked to find this 63. In frames 31 and 33 of Chapter 2. missing number: we defined two numbers whose sum (~3p3+8q2) + ( 7 ) = 0 was 0 as the or the If the sum of two numbers is of each other. a third number. then the sum less one of the numbers must equal the other number. That is. if a+b=c then c-a must ewdb. Thus. in the given problem. 0 - {-3p3+8q2) is the required number and this is 3p3- 8q2. additive inverses or’negatives 63. 64. negative or additive inverse have a sum of O -(-t+y+y2-7) or uy-y2+7 65. 66. 67. It is the negative if the sum of it and the original number is 0. 68 o ) '(6x'u'Y'9P ) ( (g) -(a+2b+7d) (e) -(-3m-12n+2q) 69. +(-6x§4yw9p) 70. (a) 8a+2a~3b+38-5b = lBa—Bb - 79 - 64. 65. 66. 67. 68. 69. 70. 71. (b) Removing the parentheses. we get -3a-[-3-4a+5]. Since 8333-8q2 when added to ~3§m equals 0. then 3p -8q is the of '31) 3248C], A number and its negative always What is the negative of -4y#y2-7? HOw can you check to see if this is the negative of -4y+y2-7? ‘Write a quantity equal to the given quantity by enclosing some terms in parentheses preceded by a - sign. (a) ~6xrhy+9p = (b) -a-2b-7d = (e) 3M+12n-2q = write a quantity equal to -6x+4y+9p by enclosing some terms in parentheses preceded by'a 4 sign. -6x&4y+9p = Remove the parentheses in the following and combine similar terms. Brackets and braces are treated the same as parentheses. (a) 8a-(-2a+3b)+(3a-5b) (b) -3a-[-3-(4a-5)] Remove only the brackets or the parentheses the first time. In the second step. remove the other. Follow this procedure when one set of parentheses is inside another. (c) xh(-x#3y-[-6xh2y]-xfiy) In each of the following. you are given three polynomials. Find the sum of all three and then subtract the result from the first one. -80- Then removing the brackets. we get -3a+3+4a-5 or a-Z. Removing the brackets first. ‘we get. -3a+3+(4a-5). Then renove the parentheses getting -3a+3+4a-5 or a-Z. Remember inside the brackets. there are two terms: '3 and (4a-5). (c) XP(-X*3y#6X#2ybxny) = x+xs3y~6xs2y+Xfiy = 3xs6y 71. (a) sum is 3y4+7y3-5y2-10y+3. 72. difference is written: (7y34-2y2-y) - (the)above This equals -3y£:$y -3 (b) sum is -21o5+3pt’c1-7103c12+3p2q3+3pq2+ difference is written: (p5-7p3q2-5p2q3>-(the above sum Removing parentheses and combining we get 3P5»31¢>"‘c1-8p‘2<.13-3m!+ . sum is ~4ab-4bo+2ac-3abc. difference is: (e) (a) 7y3+2y2-y. ~12y45-7y2. 3y”+3y-2. (b) p5-7p3q2-5p2q3- 3puq+9pqg+8p2q3: -3p5-6pqu. (c) -llab+bc-3ac. 5bc+7ab, Sac-Babc-9bc. We are now ready to consider products of monomials and polynomials. First. we must develop some of the laws of exponents. In frame 15. we defined an to mean a°a°a°... for n factors providing n is a positive integer. Thus. (p2)(p3) means (p'p)(p°p'p). This represents 5 factors of‘p and can be written as p . (muXmB) = (-llab+bc-3ac)-(the above sum) and this equals -7ab+5bc-5ac+3abc. 72° (m’m'mfim)(m'm°m) = m7 73. In general. (am)(an) = am+n if m and n are numbers. ‘what kind (Remember an is defined only if n is . See frame 15 if necessary. 73. m.and,n.must,be positive 74. In mi. m is called the integgrs as an has been defined so far only when n is a positive integer. -81.. 74. base 75. Note that the law of exponents , given in frame 72 involves the .' same base and this law holds only F when the bases are the same. x5(x9) = 75. x549 = Xlu' 76. Does x3(y2) = W57 Give a reason for your answer. 76. No. The bases are n_o_t_ the 77. To multiply par3 by psr6 we have same so the law given in to use the commutative and asso- frame 72 can't be used. ciative laws. ; Try it = 2 and y = l p2r3(p5r6) = p2p5(r3r6) by these : laws. x3(y2) = X°x'x°y°y° or there are three factors of x and two Now we can use the law (am)(an) .-= factors of y. am‘m) as we have the same base and positive exponents. xy5 = x0y-y'y°y°y or there are five factors of y and one factor Thus. p2p5(r3r6) = __ = of 3:. Since the same factors are not involved. the two cannot be equal. 77. p2"'5r3"6 = p7r9 78. (~3x2yuz)(9x3yzzu) can be multi- plied together in the same manner. Each of these quantities consists of a single term and so each is called a , 78.'monbmial 79. In multiplying monomials. we use the commutative and associative laws to change the order of the factors and to regroup the factors. Then we can use the laws of multiplication. 79. ~27x5y625 80. Multiply -2pq3r2 by ~4p2q5r2. 80. (~2)(-4)(p°p2)(q3°q5) 81. In -2pq3r2. the coefficient is . The exponent of p is 81. coefficient is -2 82. (-m np )(12np) = = 6Xponent of p is l 82- (~1)§lz)ém2)(n‘n)(p5°P) = 83. In frame 72, we stated that ~12m n2p em-en = am+n when m and n are positive integers. 83. yes because the bases are the same and the exponents are positive integers. 84. 35*7 = 312 85. 9 factors of 4 Q9 86. 28 87 No because since we don't have the same base we gannot uSe the law given in frame 72. If you're in doubt. multiply it out and compare the results. 88- <3-35>(y ~y> = 36x3y8z5 or 729x3y825 82 - Can we use this law to multiply 35 by 37? Why? 84. 35-37 = 85. The product is 312. This makes sense if we interpret the meaning 0f 35037. 35 = 3<3>(3><3)<3) or means that there are 5 factors of 3. 37 = 3<3><3x3><3><3><3> or means that there are 7 factors of 3. Hence. 35'37 means that 5 factors of 3 are multiplied by 7 factors of 3 which means that 12 factors of 3 are multiplied together. Twelve facto lg of 3 can be written as 3 In 43-46. how many factors of 4 are present? 43-46 = ZWrite using an exponent) 25(23) = 86. ZExpress using exponents) Does 73'56 = 359? Why? (3Ky7z3)(35x2y22) = 87. 88 89. 2n = (+2)“ and these both mean (_2)4 means 89. 90. 91. 92. 93. 94. 2(2)(2)(2) or +2(+2)(+2)(+2) (-2)(-2)(-2)(-2) -32 means -3(3) or -(3)(3) 0r‘-l(3)(3) and any of these equal -9 (-3)(-3) and this equals 9 a“ = -<3>(3><3><3> = -81 3)(2)(2)(2) = (a) -(3) 3 -(2 ( ~216 )( 8)= 3) ( 7) (b) (~3)(- (- ) 26 (~27>(-8 (a) 53+5 = 58 (b) (~7)“"5 = (--7)9 (0) (9)(-8) = ~72 (d) (9M) = 36 (e) -<9>- = -72 (f) -(-6u)(1) = 64 - 83 - 90. 91. 92. 93. 94. 3)(-2)(-2)(-2) = 95. _24 does not equal either of the above. -24’means the same as _(2)4 and both of these mean -1(2)(2)(2)(2) or this can be written.-2(2)(2)(2). -32 means and this equals (-3)2 means and this equals Evaluate ~34 (a) ~33<23> (b) <-3>3<-2>3 If it is possible. express each of the following results using exponents. If the result can't be expressed as a number to a power. evaluate it. (a) 53-55 (b) <-7>“<-7>5 (c) <-3>2(-2)3 (a) <-3>2<-2>2 (e) ~32<2>3 (f) -<-4>3<-1>” (e) <-2>5<-1>7 (h) s<~7>2(-%o3 What does (-5x2y)3 mean? Evaluate: -824... (g) (-32)(-l) = 32 (h) 5(49)(-§9 = - 2&3 95. (-5x2y9(-5x3y)(-5x2y) 96. -5(-5)(-5)(xz'x20x2)(y‘y-y)= -125 x9y3 or (-5)3x9y3 97. (2m9h3z)(2m“n3z)(2m”n3z) (2m9n32) this equals 16m12n12z“ 98. (m3)(m3) 99. 134(5) =3 p20 or (P5>(p5)(p5)(p5) = p5+5+5+5 = p20 ' 97. 98. 99. 100 . Complete the multiplication. (-5x2y)(-5x2y)(-5x2y) = ~______ I! (Zman3z)u means This equals . Let us see if there is a law we can use to raise a quantity to a power. Examine (m3)2. (m3)2 means In words. we could say that raising m3 to the second power is equivalent to multiplying m3 by'm . (m3)(m3) can be evaluated by the law given in frame 72. So m3(m3) = “3*3 = m6° Note that m3+3 is the same as m2(3) because 3+3 = 2(3). In other words. when a monomial is raised to a power. multiplyigg the exponents gives the same result as writing the meaning of the quantity and then doing the indicated multiplication. (p5)4 = Stated in general terms. if m and n are positive integers then (am)n = am“. mun = =

5'= _________= m -85- 100. (xu)3 = x3(ll) = x33 (m5=sh5=90>=9 101. m cube :91 or to raise x11 to the third power or x11. 102. added the exponents because am'an = m+n if m and n are positive exponents 103. Note: the bases must be the same to apply this law. or there are eleven factors of x multiplied by three factors of x which gives fourteen factors gf x which can be written x1 . 104. multiplication 105. 34(6) = 324 101. 102. 103. 104. 105. 106. (x11)3 means to do what operation? If you said that you were to multiply meaning that you were to multiply 11 by 3, you are WRONG. When exponents are multiplied, you have performed the operation of gaisigg to powers. To multiply x11 by x3, that is xll(x3) what did you do with the exponents? ‘ What gives you the right to say that xll(x3) = x11+3 = x1“? This law states that we may add the exponents when the bases are the same in order to perform the operation of <3“>6 = 25'29 = In evaluating (3p2qr3)u, we can write out the meaning of this and proceed to multiply the factors. For example: 3p2qr3) means (3p2qr3><3p2qr3><3p2qr3>. By the associative and commutative laws this can be written 3<3>(3) which equals 33(p2)3(q)3(r3)3 or 27 p6q3r9. -86.. Instead of writing out the meaning of the given statement, we can apply the law (am)n = emn when m and n are positive exponents after we have applied another law. 'We need the fact that (ab)n = anbn. By this law we can write that (x2y23)4 = . 196. (x2y23)“ = (x?)“(y)”(z3)4 107. Note that the law (ab)n = nbn only applies when n is a positive integer and when the quantity raised to the power is composed of factors. Can we write that (a2+b)3 = (a2)3+ (m3? Why? 107. NO. The law (ab)n = anbn 108. First apply the law (ab)nm a:b:mn applies gply;when the base and then apply the law (am )n is composed of factors. to find the value of The base (a2+b) is 292, (3&7b04)5. composed of factors as factors must be multiplied. Try a = 2 and b = 1. You can see the two results are pet’equal. 108. (3)5(a7)5(b)5(c4)5 = 109. Evaluate (-2p3q4)2. -35a35b5020 or 2n3a35b5c20 109. (-2)2(p3)2(q4)2= 41p6q8 110. Did you write ‘22(p3)2(qu)2 for the last problem? If you did it is wzggg, The error is in the coefficient. The correct coefficient is (-2)2 as an exponent affects the letter. number or symbol which immediately precedes it and in this case. the exponent 2 affects everything in the parentheses, and the -2 is inside the arentheses. Remember --22 and (- 32 do not mean the same and are 9.01 equal. - 87 - Evaluate (-3x2)”. 110. (-3)“(x.2)4 = 81x8 lll. Evaluate: (a) (.5p3q6r5)2 (b) (~a3b2c5>6 (c) (-y“z8>“ 111. (a) ( 5)Z(p3)2(q6)2(r5)2 112. Evaluate -(3nnz)2(-2n3n‘*). 25p6q12r10 Write out the meaning of this (b) (-1)6(a3)6(h2)6(e5)6 statement before you try to simplify. a18b12e3° (c) (-1>““<28>4 = y16232 Remember in parts b and c that -a means -l(a). 112. -(3mn2)(3mn2)(-2m3nu) or 113. Instead of writinfi)out the meaning of -(3mn2)2(-2m3n we could (-l)(3mn2)(3mn2)(-2m3nn) square 3mn2 and then perform the multiplication. These 9 (-l)(3)?3)(-2)(m m°m3)(n2n 2 n4) Doing this. ~(3mn2)2(-2m3n“) = or 18m5n8 ~(9m2nu)(-Zm3n#) and this is <-1><9)<-2) or 18m5n8. -(-p2q)3(-3pq2)2 = 113. -(-p6q3)(9p2q4) = 9p8q7 114. Perform the following operations and simplify. (a) (35)” Leave in exponent form. (b) [(-2)3]5 Leave in exponent form. (c) (5)2(53)2 Leave in exponent forM. (d) (34p3q5r>2(392qr5)3 (a) -<-2xy3>7(-x3y“)3 114. 115. 116. 11?. 118. 119. -88.. (a) 320 (b) (~2)l5 (c) 52°56 = 58 (d) (38p6q10r2)(33p6q3r15) = Bunuanl" (e) -<[-2]7x7y21><- 9y”) = -(-128x7y21)(-x9y12) = ~128x;6y33 115. by the distributive law 116. 3a+3a(3a> 117. 3a3+9a2 118. -2p3(3p2>+(-2p3)(7p) = 119. -6p 5-1“,th You may have written -6p5+ ~13p4 as your answer. This is correct. However. we generally use only one sign between terms since adding a negative is the same as subtracting a positive. we can write -6p5-l#p4. -X3y2(4x3)-X3y2(-3xy) -X3y2(5y3) = -4x5y2+BX&y3 -5x3y5 120. So far we have multiplied together quantities which consisted of factors. Now let us consider multiplication when terms are involved. 'we multiplied 3(4+7) by two different methods. In one case, we said 3(h47) = 3(11) = 33. In the other case, we said 3(4+7) = 3°4+3‘7 Why can we make this last state- ment? The distributive law states that a(b+c) = ab + ac providing a, b and c are rational numbers. Using the distributive law. 3a = Complete your answer to the last frame. -2p3<3p2+7p) = The distributive law can be extended to read a(b+c+d) = ab+ac+ad+ae a(b+c+d+e) = ab+ac+ad+ae a(b+c+d+...+n) ab+ac+ad+...+an Using the distributive law. ~x3y2(4x2-3xy#5y3) = 9mn3(3m2n+4mn3-m3) = 120. 121. 122. 123. 12a. - 89 - 9mn3(3m2n)+9nn3(umn3)+ 121. 9mn3(-m3) = 27m3nu+36m2n6 -9m4n3 ep6q2+3p5q3-6p5q3-2p4q3 by 122. the distributive law. Combining similar terms, we have -p6q2-3p5q3-2p4q3 (a) -6x:3y52ll you don't need to use the distributive law here as only factors are involved. 123. (b) Use the distributive law here. ~12x3w3+21x2w5-15xw7+27xw3 Use the law (ab)n = anbn to get <-2>‘*“4‘*. Now use that (am)n = amn and get l6p20q8wl+ Use the distributive law. -3x“y2-3x3y3-8x3y3+8y5 which equals -3x?y2-llx3y3+8y5 (e) (d) (2a+3)b+(2a+3)3o = 124. 2ab+3b+6ac+9c (Ba-2)ha+(3a-2)l = 125. 12a2-8a+3a-2 = 12a2-5a-2 -pq2(p5-3puq)+2p3q(-3p2q2-pq2) = Simplify the following. (a) (2m3z5)(-3x2y226) (b) -3m3<2+x2-7xw2+5w‘*-9> (o) <-2p5q2w>“ (d) ~3xy2(x3+x2y)+2y3(-4x3+4y2) The distributive law can also be extended to (a+b)(c+d) Let (a+b) = N. then (afib)(c+d) becomes N(c+d). By the distri- butive law N(c+d) = Nc+Nd. Since N = a+b, Nc+Nd becomes (a4b)c + (a+b)d which equals ac+bc+ad+bd. Using the distributive law, (2a+3)(b+3c) = (3a-2)(ua+1) = = (3p-2q)(p2+#pq-3q2) = 125. 126. 127. 128. 129. 130 131. 132, - 90 - (3p-2qxp2+(3p-2q)4pq+ 126. Since multiplication is commutative (3p-2q)(p2+4pq23q2) = (3p-2q)(-3q2) = 3p3-2qp2 (p2 +apq.3q2 )(3p-2q>. ‘+12p2q-8pq2-9pq2+6q3 = By the distributive law. (p2+“pq-3 Sq §)(3p-23) equals 393+10p2q-17pq2+6q3 ( M2+4pq-3q )3p+( p +“pq-3q2 )(-2q). When this is expanded and similar terms are combined, we get the same result as in frame 124. Ybu can use either expansion. (st3y)(x3-ax?ya5xy2+3y3) = 5x(x3-4x?yb5xy2+3y3) 127. (In-2n)2 means ~3y3 = 6x2+5xs6 (c) (2y-3)(2y-3) = uyz-lzm (a) <9p446p2q3+nq6>3p2+ (9p4i6p2q3+“q6)(-2q3) 27p6-8q9 (e) (7a4e8o3)7a4+(7a4+8o3) (~8b3) = 49a8-6uh5 factor 1350 a2+2a 136. factor m3 138. 32 because a3'a2 = a5 139. (a) (4a-3b)(a2-ab-2b2) (b) (Bx-ZXZX-i-B) (c) (2y-3)2 (d) <9p446p2q3+4q6><3p2-2q3> (e) (7a448b3)(7a“e8b3) If (3xp2)(2x+3) = 6x?+5xs6. then we may say that (3xp2) is a of 6x2+5x+6. a(a+2) = . a is a of a2+2a. Name another factor of a2+2a. If m is one factor of m“. the other factor is If a3 is one factor of a5. then the other factor is because Since a3°a2 = a5. then gé_= a3 139 . 140. 1le. . 142, ln3. 1M4. .. 92 .. a2 1%. y5 because y7°y5 = yl2 141. p9 142 . a%3 Remember any numb er divided by itself is 1, so ll+3. 1144. because the exponent in 145. the numerator isn't larger than the exponent in the denominator. If we use the law given in frame 140. we get :63. what does this mean? At this POint. we don't know as x'3 hasn’t been defined yet. I2 = because . y? In general terms. if m and n are positive integers and if m>n, then P am -n . ”:3 P’s To divide £12 , we consider this x3y2 as 3:1 . i and now we can apply x3 y2 the law given in frame 1&0. To divide x2 by x5 that is _x_2_ . 5 we cannot use the law stated in frame 1%. Why can't we use this law? We will write another expression which is equal to the given one. x2 x5 The numerator and denominator have two factors of x in common. Or we may state that x(x) is a common factor of both the numerator and denominator. ‘x O O O O XXXXX 14 . 5 :jbecauseifi (3,9) =y5 " O‘IF’ K0 148- 334-4 (This goes in both blanks) - 93 _ 1146. 147 . 148 . lib9 . Dividing both numerator and denominator by the common factor. we have X'X + x'x _ l __ ac'x'x'x'xo- X°x " x'X°x - l 5 . Thus, it: = _l_.__ because -]-'-(x5) = x2 x5 x3 x3 Z; = because . y In more general terms. if m and n are positive integers and m= 9633378251332“ lingyw-l6x-l digided 56xyyw64x3-hx2 by z. 160. Divide 3a2-5a-2 by a-2 161. descending 159. 160. 161. 162. If 4x2y3 is one factor of 64x3y3-100x2y6-hflzy3. the other factor is . Check your result. H-6x3y2 is one factor of H96xy3+78x5y2-3x y , the other factor is . Check your result. If 4x2 is one factor of 56x9y+64x3-4x2. what is the other factor? ‘What operation did you use to obtain this factor? If a-2 is one factor of Bag-5a-2. how would you obtain the other factor? When you are to divide a poly- nomial which consists of more than one term, the division is set up as a long division problem. Both 0 omials must be arran ed in the same order. That is, both polynomials must be in descending order or else both.polynomials must be in ascending order. Both 3a2-5a-2 and a-2 are in order. The division problem would look like this: a-Z / 3a2-5a-2 Now divide the first terms of each polynomial. '-#- 1623. 3a 163. Multiply 3a by a-2 a-z @472 gag-6a 161+. subtract a a-z 3a2-5a-2 3a2-6a +a 165. brought down the next number or brought down the 2. 163. 164. 165. 166. ~96- In this case. divide 3a2 by a. so the first term in the quotient is 3a The problem now is: a-2 3a -5a-2 This is like long division in arithmetic. If we were to divide 542 by 36, we would estimate the 1 quotient, 36 / 542 , and then multiply the partial quotient by the number we are dividing by and then find the difference of these numbers. 1 36 / 542 36. 18 we do exactly the same process when dividing polynomials. a In 3-2 / 3aZ-5a-2 , what will we do with the da? Our problem will then be ? The next thing to do is , Do this. 1 In 36 / 542 , you next . i3- YOu do the same thing when dividing polynomials. a a-Z I 3a2-5a-2 gag-6a a-2 You then repeated the above process until you had no more numbers to bring down. Look at the above problem. You will now . (If necessary look at frames 161-165.) 166. 167 . 168. 169 . 170. 171. 172. divide a by a 1 a+ 1 a-Z / 3a2-5a-2 gag-6a + a-Z 1:2. 0 arrange both.polynomials in the same order. (See frame 160.) 2a+5 / 6a3+llaZ-2a-8 6a3 by 2a :multiply’2a+5 by this result Subtract jag-2a+4 2a+5 / 6a3+lla2-2a-8 6a3115a2 - 4a2-2a - 4a2-10a 8a- 8 8a+gQ -28 - 97 - 167 e 168. 169. 170. 171. 172. 173. The quotient is . Complete the division. The remainder in this division is . Since the remainder is O. we can say that a-2 is a factor of BaZ‘Sa-Zo In order to divide llaZ-Za-B-I-éa3 by 2a+5. we must first Set up the division arranging both polynomials in descending order. We first divide . and then . Then we , and bring down the next number. This process is repeated until the division is completed. Do the division: 2a+5 / 6a3+11a2-2a-8 ‘We write the remainder just as we did in arithmetic. The remainder here is -28 and is written -38 . 2a+5 In other words. 6a3+lla2-2a-8 divided by 2a+5 equals 3a2-2a+4+ .:ZQ. 2a+5 Is 2a+5 a factor of 6a3+lla2-2a-8? Give a reason. 173 e 171%. 175. 176. .. 98 .. No because there is a 174. remainder. (See frame 168.). Multiply 2a+5 by 3a2-2a+4 175. and then add_~28 to the product. This should equal 6a3+11a2-2a-8 (2a+5)(3a2-2a+4) = 176. 6a3+11a2-2a+20 This + -28 = 6a3+11a2-2a-8 which is the required quantity. How would you check this problem? Do the check. Divide 3a3-l6-12a-I-4a2 by 3a+4. Check your result. Arrange in order first. 177. When you found the difference a2 -4 3a+4 / 3a3+4a2-12a-16 3a314a2 ~12a-16 ~1ga-16 Check: (a2-4)(3a+4) = 3a3+4a2-12e-16 177. Mange in grder first. 178. 178 e .3a2-7a-6 2a-5 / 6a3-29a24-23a4-34 a3- a2 2 -l4a +23a -l4a213§a ~12a+34 -lg§130 4 2- _ 4 3a 7a 633:3, 179. between 3a3+4a2 and the product of a2 and 3a+4. the difference was 0. In this case you have to bring down more than one number as you would in the case of dividing 3742 by 18. ..Z.. 18 / 3742 14 Here 14 is not divisible by 18, so a 0 goes in the quotient and the next number is brought down. 20 18 3742 39 142 Divide: 6a2+23a+34-29a2 by 2a-5 'Write the complete quotient of the last problem including the remainder. Quotient is , Is 2a-5 a factor of 6a3+23a+34-29a2? Why? -99.. 179. No because there is a remainder. 180. 18p3-9p2q-32pq2+16q3 181. 181. Divide to find out. Be 182. sure that both.polynomials are arranged in the same 180. If a polynomial contains more than one variable, consider only'gng, variable when arranging it in order. For example, to arrange 18p3-32pq2416q3.9p2 in descending order. choose one of the variables and arrange the polynomial in descending order for that variable. Choosing the variable p and arranging the polynomial in descending order. we get Is 3p+4q a factor of 19p3-32pq2+16q3-9p2q? In the polynomial a3-4a+1, the a2 term is missing. order. It is often necessary to consider missing terms when we are 622 llpqii.4q2 dividing. 3‘qu / 18p3-9p23-32pq24-16q3 18231242 9 The coefficient of any missing -33p§q-32pq§_ tam ”'5 - -33p q-44ng_ 12pq2+16q3 1&9211693 Yes it is a factor. 182. O 183. Write the polynomial a3-4a4-l 183. a3+0'a2-4aA-l 184. 184, a2-e-3 185. a+1 / a3+0°a2-4a+l a3: a2 - a -4a - a - a ~3a+l ‘33- 341' Quotient is aZ-a-3+-HL- a+l putting in the a2 term. Divide a3+0’a2-4a+l by a+1. You could divide a3+0'a2-4a+1 by a+1 by arranging both polynomials in ascending order. The division then would be 1+a / 1-4a+0°a2+a3 Do this division. 185. 186. -100- l-5a+5a2 l+a / 1-4a+0~a2+a3 11 a ~5a+0oa2 -§a-§a2 5a: + a; is +§a -4a3 Quotient is l-5a+5a2 4. .2433 l+a From frame 183. (a+1)(a2-a-3) = a3-4a-3 Adding 4 to this we get a3-4a+1 which is the 'required quantity. From frame 184. (1+a)(1-5a+5a2) = l-4a+5a3 Adding this to -4a3 we get the reguired quantity of l-4a+a 186. 1870 In frames 183 and 184. we've divided the same polynomials and gotten different results. However. both will check. Check the results for frames 183 and 184. You will get different quotients if the polynomials are arranged in different orders when one polynomial is ngt_a factor of the other polynomial. Both results will check and thus both quotients are correct. If you are to divide 5x3-x2+3 by x22, it is more desirable to arrange both polynomials in descending order. as then 5x3 divided by x gives a number which doesn't involve a fraction. If ascending order is used. the problem is 2+x I 3+0°th2+5x3 and the first step involves dividing 3 by 2. The result involves a fraction which then must be gultiplied by 2+x. 1 2 2+x /'3 + O'x - x2 + 5x3 1 34-15}: l 2 - 15x - x You can see that the rest of the problem will involve fractions as now we must divide -1%x by 2. Do the division, first arranging in descending order and then arranging in ascending order. Check both results. ~101- 187 . 3 188. Divide: ea- W (a) Rah-10a3b4-4a2b2-ab3 by Za-b x22 / 5x3- x? + O'x:+ 3 (b) 4p-2p3+21.7p2 by M Egig§25x?-llx#22) + —41 = (c) th-2p2q2-39u*1093q+5pq3 by 5x3'x2*3- qZ-p2+3pq z+x./ 3+0 x - x3 + 5x3 (d) 64m6-27q3 by'4m2-3q Quotient here is l 2 ‘ £2 ‘ %x3 * igfg affine-gee) + 52cm 5x3-x2+3 188. (a) 6a3-2a2b-Ivab2 189. Is w-Z a factor of w3-8? Why? (b) ~2pZ-p+7 (c) Zqzépq+392 (d) 16m4'4-12m2q‘0-9q2 There are several missing terms here. and these all have coefficients of 0. You can supply as many missing terms as you need. 189. Yes because when w3-8 is 190. Write 113-8 as a product. vided by w-Z. we get +2w+4 and there isn't any remainder. 190. (w-2)(w2+2w~I-4) 191. Determine whether the first polynomial is a factor of the Remember if g- = c then second polynomial in each of be = a. the following. (a) 3k+2; 27k3+8+36k+54k2 (b) 16a"-12a2b2+9b"; 64a6+27b6 (c) 2x-3; 4x3-l7x410 191 . 192. 193. - 102 - (a) yes. quotient is 192. 9k2+12k+4 (b) yes quotient is 4a2;3b2 (c) no. quotient is .221. 2x2+31~4+ 2x23 (d) yes. quotient is 9.2-43.2 (e) no. quotient is -142c+5cz+ -—39—-—- 7-4c+3c2 Note: If the polynomials are both in ascending order in part 6. you get the above result. If you used descending order. you have a different quotient. Check it to see if it is correct. (a) (3k+2)(9k2+12.k+4) 193. (b ) (4a2alv3b2 ) (16:14-12 a2b2+9b’*) (a) (2x-51)(9x2-412) (a) (4x?+20x#25)-%194. (ha2-20+25)= (b) (6Z1P 3-11a2+20a-8)x .- 4+2a3 -2a &a+10) = -a 4+4a 3-9a2+19a-18 ( -48 a- - = °)§§32.47;2§Hy2 ye) Note: M must be done by y2+y+l long division (d) eriy: 18x3-Bxy2-45x3y+20y3 (e) 7-40+3c2: 24c2+18c-7+l8c“;14c3 write the seggnd polynomials of parts a. b. and d of frame 190 as products. Perform the indicated operations and simplify. (a) (2x.5)2-(2x.5)2 (b) (3a2-4a+8)(2a—1)-(a3-2a+5)(a+2) (C) (337.8)?- .. 1328:; y2.y..1 Remember -- if the result of an Operation is a polynomial which contains more than one term and this result is then to be sub- tracted from another’polynomial. enclose this result in parentheses so that you will remember to sub- tract the entire quantity. - 103 - 194. (a) -x9y32 (b) ~3m315(8m3n12) = 24mZy5n12 (c) -12p5qr6p4q“+ §p3q6 (d) 16p2-2'tpq’r9q2 (e) 3b3-15b2+21b-3-(2b3+7b2-4b) = b3-22b24-25b-3 (r ) .8m2p24-1-1Lan5p5 (g) Ybu need to use long division here. Be sure to arrange in order first. Quotient is 5x21. For example: (2x#5)2-(2x-5)2 = 4x2420x425- (4x2-20x+2 5) or 412+20x+25-4x2+ng—g§ It 99.2213 equal 41c2+ZOx+2 5-4x2-20x+2 5 Perform the indicated operations and simplify. (a) (23153 (b) - ”flm‘w (c) 3p3q(-4p2-2pq34%q5) (a) (up-3oz (e) 3(b3-5b2+7b-1>-<2b2-b>(b44) (f) 56m393-2m32128m726 -7m2p (g) 15x3-42x:§:2x2 3x2+2x+8 Chapter 4 - Factoring In this chapter, we shall learn to factor certain kinds of poly- nomials. The problem of factoring is to write a polynomial as the product of polynomials. 1. In Chapter 3.5. frame 151}, you were told that 3p q was one factor of 27p3q2-126p2q5-3p2q. How would you find the other factor? 1. Divide 27p3q2-126p2q5-3p2q 2. If polynomials can be expressed as the product of two or more factors. by -3p2q then factors are concerned with what Operation 2. multiplication 3. If -3pzq is one factor of 27p3q2-126p2q5-3p2q. the other factor is . 3. -9pq+l&2q3+l 1J1. 27p3q2-126p2q5-3p2q expressed as a a product is . 1*. -3p2q(-9pq+1+2q3+l) 5. If 4:: is one factor of lug-8x, express Luce-8x as a product. uxz—ax = 5. Macho-2) 6. 4:: is called a common £3.53; of Lug-8x. Each term in 1+ -8x can be divided by 4:: and no fractions are obtained in the result. 1+): is called the ar as common m of lug-8x. If a polynomial has a common factor. the polynomial can be expressed as the product of a monomial and another polynomial. If the monomial is the m common factor. then this second polynomial is irreducible to a monomial multiplied by another polynomial o What is the largest common factor of 9y2-6y? - 101+- 7. 8. -105... 3y(3y-2) 27X? 9 o 27W(X-2) 10. Either 7pq2 or ~7PC12 7p q2(~6p «1210127) 9m2(3m-6m3+7 ) O r -9m2(-3m-6m3-7) 13. (a) 16m3(-m244-3m) or (b) -16m3(m2-lH-3m) If only one possible set of factors is given. it doesn't mean that your result is incorrect. Multiply and check to see if your result is also correct. 141w3(1-5x2y2+2w) Be sure that you have 3 terms in the one factor. 7. 8. 9. 10. 14. Express 9y2-6y as a product. Since 3y-2 is not reducible to the product of a monomial and a poly- nomial, we say 3y is the largest 92.132191 gagmr of 9y2-6y. What is the largest common factor of 27x2y-5uxy? Express Z7xzy-54xy in factored form. Note that when 27xy(x-2) is expanded. we get 27x2y-54xy or the same thing we started with. What is the largest common factor of -u2p2q3+84p5q2449pq2 ? Factor 42p2q3+84p5q2+49pq2 using 7pq2 as one of the factors. Factor 27m3-54m5't63m2 . Note that in the second factor there are the same number of terms as in the original polynomial. 27m3- 541115463012 has terms. The common factor of 9m must divide each of those terms and so we get 3, terms in the other factor. Factor: (a) -16m5+61+m3-’48m’+n (b) lhm3-7Ox3y54-28x2’y“ To factor 3(a-b)+a(a-b) we use the same technique. Is there a common factor for both terms in 3(a-b )+a(a-b)? Remember () are used to consider a group of terms as one number. What is it? 14. yes there is a factor which divides both terms. It is (a-b). 106 - 15. 15. divide 3(a-b)+a(a-b) by (a-b). 16. 16. a-b a a-b - fat—b; "' {a-bg " 3 " a 17. l8. 19. 20. 21. 22. 23. So (3+a) is the other factor. See if (a-b)(34-a) equals 3(a-b)+a(a-b). or multiply (a-b) by (34a) and multiply 3(a-b)+a(a-b) and see if the results are the same. (ac-1+3?) (x-fiy)(2x-3y) only by 2:: by (2x-3y) No. Only (x-’+yM2x-3y) is in fa(”sored form. Common factor is (Bx-Y). (Bx-y) (zxwy-Bz ) 170 18. 19. 20. 21. 22. 23. 21+. Then a common factor of 3(a-b)+a(a-b) is (a-b). If there is no other number which divides both terms this is the largest common factor. To get the other factor. we must what Operation a-b +a a—b gives what other a-b factor? Expressed as a product. 3(a-b)+a(a-b) = (a-b)(3+a). How would you check to see if this is correct? Be specific. What is the largest common factor of 2x(x-4y)-3y(x-’+y)7 Factor 2x( x-‘+y)-3Y(X-“Y) . If you wrote (x-lbykx-By. you are 2229.3. Consider (it-#ykx—By, how much is (ac-4y) to be multiplied by? In (x-QyXZx-By), how much is (x-hy) to be multiplied by? Obviously then (xr4y)2xh3y and (xaQy)(2x~3y) can't equal the same quantity. Are both of these quantities expressed as factors? Which one(s) is expressed in factored form? Factor 2x( 3x-y )flfl 3x-y )-32( 3X-y ) . Factor 310(p4-q )+(P+q) . 21+. 25. 26.. 27. common factor is (a+b). 28 . (ab-ac )+(xb-cx) or -107- The common factor here is (m). (mXBpfl) Remember +(p+q) = +l(p+q). 25. 3(a-b-c)(X43y) 26. If you had (a-b-c)(3x+9y). note that 3x+9y has a common factor of 3 and you should factor again. If you had (a-b-c)3(x+3y). is your result the same as the above? mm (a) 3x?y(x3-y2><9xy+3y“-1> If you had (la-5'2) (27x3y24-9xzy5- 3x230 you need to factor again. 0)) (12+3X)(4x2+8xy) == X(x+3)4X(x42y) = ll-x2(x¢3) (x+2y) 27. 28. c(a+b)+d(a+b) = (a+b)(o+d). 29. (9.be )+(-ac-cx) Factor 3x( a-b-c )+9y( a-b-c ) . Be sure that there are no common factors in any of the factors of your result. Factor completely. (a) 27x3y2(x3-y2>+9x%y5- 3x2y(x3-y2> (b) h>r2(x2+3x)+8xy(x2+3x) In the polynomial ac+bc+ad+bd. there is no number distributed over the entire sum. Hewever. notice that c is distributed over the first two terms and d is distributed over the last two terms. Using parentheses to group the first two terms and the last two terms and kegping the same value we have: ac+bc+ad+bd -.= (am-be) + (ad+bd). This can be rewritten as c(a+b)+d(a+b). The terms here have a common factor of . Factor c(a+b )+d(a+b ) . So even if we don't have a number distributed over the entire quantity. we can sometimes factor the polynomial providing we group the terms so that the resulting terms have a common factor. Group the expression ab-ac4-xb-cx so that each group has a common factor and the value is equal to that of the given quantity. If you didn't look at the answer to the last frame look now. Notice that there are several ways of or (ab+xb)-(ac+cx) 29- a(b-c)4-x(b-c) = (b-c)(a+x) 30. yes because multiplication is commutative. 3].. there is n9.common factor. 32° (PZ-pq)+(qr-pr) = PCP-q)-r(-q+P) = P-q)(p-r) N°te (p-q) and (-q+p) are t 9 same. 33. yes - 108 - 30. 31. 32. 33. 34. grouping the terms so that each group has a common factor and the entire quantity equals the original one. Factor: ab-ac4xb-cx (ab-ac)+ (bx-ox) :- Could we have written (a+x)(b-c) instead of (b-c)(a+x)? Why? Suppose we had group ab-ac+xb-cx as (ab+xb)+(-ac-cx). The first group of terms has a common factor of b and consider c as the common factor of the second group. We then would have: (ab+Xb)+(-ac-cx) = b(a+x)+c(-a-x). What is the common factor in b(a+x)+c(-a-x)? Thus. we cannot factor b(a+x)+c (-a-x) as there is no common factor. However. we could consider -c as the common factor of the second group and then we have: (ab+xb)+(-ac-cx) = b(a+x) - c(a+x). This has a common factor of (a4x) so we may write ab-ac+xb-cx.= b(a+x)-c(a+x) = (a+x)(b-c). 2 Factor p ~pq+qr-pr. Show all steps. There are two conditions which must always be met. The quantities must be equal and there must be a common factor. You were given pz-pq-l-qr-pr. IS p(p-q)+r(q-p) equal to this? Therefore the first condition is met. Is there a common factor in p(p-q)+r(q-p)? 37’. - 109 1K) 35. No. 36- JFactors are connected by multiplication. <3a2-3ab>+<-uac+ubc> = 37. 3a(a-b)-4c(a-b) = (a-b)(3a-4c) (2xy-4wz)+(-3wy#6wz) = 2x(y-Zz )-3W(y-ZZ ) = (yrZZ)(2xr3W) 380 (nyahxz)-(3wy+6wz) doesn't equal the other quantity. (2xy-flwz)-(3wy+6wz) equals Zia—“wa-Bw-éua 39. 39- Ea) 3(p+q)+a(p+q)=(p+q)(3+a) 40. lmy(x+2y)+3ab(x+2y) = (xv-Zyxhxyflab) (c) 2x7~(x-y)+5y(x-y)= (x-y) (22:24-53?) (d) 3a(a;b)-hz(a-b)=(a-b)(3a-4z) (9) 2p(2Mp)-3q(Zr-p) There is g2,common factor here so try another grouping of terms. It happens in this case. that pg,grouping works, so this gannot be factored. Then the second condition is n93 met. Is p(p-q)+r(q-p) in factored form? Why? Factor 3a2-3ab-4ac+4bc. If you had 3a(a-b)+4c(-a+b). this equals the original but can't be factored as there is no common factor. Factor 2xyb4xz-3wy46wz. Suppose you had written that 2xy-4xz-3wyoéwz = (2xy-4xz)- (mm-6%). What is wrong with this? Factor the following. Be sure that you remove §;;_the common factors. (a) 3p+3q+ap+aq (b) 4xzy1-8xy243abx-I-6aby (c) 2x3-2x2y4-5m-5y2 (d) 3a2-3ab-ha244bz (e) 4pN2p2-6qr‘fi3pq (f) 6p2r-3p3fl2pqr-6p2q (g) up3-8p2+p-2 (h) 9x3-27x2-x4-3 (i) 2p3+2r-4p2-pr If 4x2 is one factor of 56x9y*64x3-hx2. the other factor is . What Operation did you use to obtain this factor? MJL. 4&3. 1+3. 45. - 110 (f) 2p2(2rsp)+6pq(2r~p)= (Zr-p)(2p2+6 pa): (Zr-p)2p (p+3q ) 3r2%p(§§'§3( >32 2)(421) p2 p- + p-2 r p- p + 9x2(xe3)-l(xr3)=(xP3)(9x2 -1) In order to group and get terms having a common factor. you have to consider this as (g) (h) (i) (2p3-4p2)+(5pr+2r) or as (2P3rpr)+(-#p2 +2r). (2g 3-4p2)+(-pr+2r)= 2p (p-2 )-r(p-2) (P-Z) (2p 2-r) 14x2y+16x~l divided 56xuy-r64x3-lex2by 1+x2 divide Bag-5a-2 by a-Z 41. 42. 3a+l 43. multiply 3a+l by a-2 2, 45. is 46. is not If a-Z is one factor of 3a2-5a-2. how would you obtain the other factor? If a-Z is one factor of 3a2-5a-2. what is the other factor? How would you check to see if this is correct? We multiply 3a+l by a-Z using the distributive law. (3a+l)(a-2) = 3a(a-2)+l(a-2) Each factor contains hOW'many terms? A polynomial containing two terms is called a EEEQEEELo 3xy-hx2 a binomial. 2a3-3ab24-7b3 a binomial. The product of two binomials occurs very frequently in algebra and therefore it is useful to use a shortcut in order to get the product more quickly. However, remember if you forget the short- cut. you can always use the distributive law. lHS.. 6ax +20y2 4?. -8ad +15ab -111- 47. In (a-b)(2a+b) = 2a2-ab4b2, notice that the first term in the product. 2a2, is the product of the first tenms in each binomial. Next notice that the last term in the product. -b2. is the product of the last terms in each binomial. In (3xehy)(Za-5Y). the first tann in the product is _______ and the last term in the product is Expanding (3xe4y)(2a.5y) by the distributive law. we get 3x(2a-5y)-4y(2a- y) = 6ax915xyb8ay+20 . So far by the shortcut. we have (Bx-4y)(2a-5y)=6ax _________+20y2 'We are missing -l5xy~8ay. Notice that -15xy is the product of the first term of the first binomial and the second term of the second binomial. Also notice that -8ay is the product of the last term of the first binomial and the first term of the second binomial. Thus to multiply (2&43b)(5a-4d). by the shortcut. we would first multiply 2a by 5a getting lOaZ. Then we multiply the first term of the first binomial by the last term of the second binomial getting . Then we multiply the last terms of each binomial getting -12bd. Thus (2a+3b)(5a-l+d) = lOa2-8ad+15ab .. bd. If we expanded (2a+3b)(5a-hd) by the distributive law we would get the same result. - 112 - 48 . ~10ac+2b c+15ad- 3bd #£?.. hayimgyZ-lZXZyZ-Byn 'We now have some similar terms which we can combine getting 4x9-llx2y2-3y4 50- (a) 6p2-l5pqeéap-15aq (b) 35n2-20n+14m-8 = 35m2-6m-8 1+9. 50. 51. (c ) 12p 2+2np+27p+ 514:12p24 5lp+51+ (d) Write this as (5x~3y)(5§-3Y) product is 25x2—30xyo9y (e) aZ-ab-o-ab-bZ = aZ-bZ (1‘) Write this as (a-b)§a-b). product is aZ-zab+b (a) 9m2+12mn-12mn-16n2 (-20*3d)(5a-b) = The product of the first term of the first binomial and the last term of the second binomial and the product of the last term of the first binomial and the first term of the second binomial are referred to as ggggggpggdgggg. (xz-Bythxztyz) -- When the binomials have similar terms. we shall always have similar terms in the product which we can combine. Whether we multiply using the distributive law or by the short- cut. notice that each term of the first binomial is multiplied by each term of the second binomial. Remember this shortcut only applies when multiplying two binomials. Find the following products. (a) (Zp-Sq)(3p+3a) (b) (7m-4)(5m+2) (e) (4p49)(3p+6) (d) (ix-3y)2 (e) (a-b)(a+b) (r) (a—b)2 (g) (3m-4n)(3m+#n) <5x2-6y) In parts 9. g and h of the last frame. the product contained only two terms. That is the two cross products had the same absolute values. Since these terms had different signs. their sum.was O. This will always happen if the two = 9m2-l6n2 binomials are the sum and difference of the same two numbers. - 113 _ (h) 2538+.30X2y4130x2y- 36y2 = 25x9e36y2 51. (a) lémZ-ZSP2 52. (b) 9e2+3eb-2b2 fig carefg; here. This isn't the sum of two numbers multiplied by the difference of the same two numbers. 52. 9m6-25n“ 53. 53. (a) 161311-qu10 51., _1_ a___ 4 (b) 4" 113‘ (c) 9p2-9pq3-4q6 Be careful here. This isn't the sum and difference of the same two numbers. (d) 9b2-24bc41602 Be careful here too. 54. factors 55.8 For example. (a-b)(a¥b) is the difference of a and b multiplied by the sum of a and b. so the product will be a2-b2. (a) (“n+5p)(hm-5p) = (b) (3a+2b)(3a-b) = Let us consider the product of the sum and difference of the number “m and 5p. (um+5p)(nm-5p) = 161112.251:2 16m2 is the square of hm and 25p2 is the square of 5p. In other words. the first term of the product is the square of the first term of either binomial and the last term of the product is the square of the last term of either binomial. The product is the difference of these squares. (m3-5n2)(m3+5n2) = Complete the following. (a) (#p6+3q5)(4p6-3q5) (b) (gach-%yZX%-:<2+%y2) (e) (3p+q3)(3p-4q3) (d) (3b-#0)(3b-4c) Since <3m3-5n2><3m3+sn2> = 9m6-2snt we can say that 3m3-5n2 and 3m3+5n2 are of 9m6-25n4. 05 if we are asked to factor 9m -25n4 . or to write 9m6-25n4 as a product. we can write 9m6-25n“ = (m3-5n2)(3m3+5n2>. - 114 - The expression 9m6-25n4 is the difference of two squares. The factors of this expression are the sum and difference of the square roots. The only thing now is to be able to determine the square root of a number. The square root of a number is 55. a number which when 56. In symbols, Vi? = a m if a(a)=N. Find the square root of: (a) 64 (b) 16 (c)m8 (a) 25106 (e) 9p2q10r6 56. (a) 8 57. Sometimes you will wish to find (b) u the square root of a number such (c) mu as ZBOM. There is an arithmetic (d) 5p3 process by which this can be done. (e) 3pq5r3 However. we will factor the number and then make use of the defini- tion given in frame 56. Factor 2304. Choose any|number which will divide 230k. (A number which divides another gives a quotient with no remainder.) 2304 is an even number so 2 must divide it. 2304 = 2’1152 Continue this process and factor 230# until you get a number which you recognize as a perfect square. 230“ = 2‘ 57- 2304 = 7322-2-le 58. The definition in frame 56 states or W = a if a'a = N. Therefore. in 2301» = 2-2-2-2-2-2-36 2020:2144, 2‘2 must represent a perfect square whose square root is 2. So. V(2-2)(2°2)(1MT = 2212 is 48 Use this to find V2-2'2-2-2-2°36 58. V(2°2)(2°2)(2°2)(36) = 2°2'2'6 = #8 59. (a)V—-_2-2~81 = 2.9 = 18 (b)\/2°2‘2’2’49 = 2°2°7 = 28 (o) V2°2°2°2°121 = 2-2-11 = (d)\/3‘3°3°3‘49 = 3°3'7 63 60. 5x2 because 5x2(5x2) = 25X“ ByA'because 8y#(8y4) = 64y8 61- <5x2+8y4><5x2-8y4) - 115 - 59. 60. 44 61. 62. Find the square root of: (a) 324 (b) 784 (c) 1936 (d) 3969 Let us once more consider fa toring polynomials such as 25x9-64 . In frame 52. we said that if we multiplied tagether the sum of two numbers and the difference of the same two numbers the product would consist of the difference of the squares of these two numbers. For example, (xF3)(x+3) = x2-9 Thus we can say that if we have a polynomial which is the difference of two squares. then its factors are the sum and difference of the square roots. 1e. the factors of xz-lé are (xs4 and (xw4), where x is the square root of x? and 4 is the square root of 16. For ex To factor 25x9-64y8. we must first determine if both of these terms are squares. V2523 = . Why? V64;B = . Why? Then the factors of 251:.“2-6437'8 will be the sum and difference of the square roots. Factor 25x4-64y8. Factor each of the following. Be sure and check to see if each of the terms is a square. (a) w2-49 (b) 4x2-81y2 (c) 9p6-16q12 62. 63. -116- 83 EM???) > 2x» ZXh (c) 3p3+4qg)(3p§zuq6) (d) This can‘t be factored as it isn't the difference of two squares. 12 isn't a s uare. (eg (xv -3)(xy3+3) (f This can't be factored. Both terms are squares but we don't have the difference of them. Note: Some of these non- factorable polynomials may be factorable at a later time. However. they are not factorable by the rules we've had. Factoring this as the difference of two squares. we get (ue5+16h“)(ue5-lob4) 64. 4 650 yes. 63. 64. 65. 66. (d) m6-12 (e) x?y5-9 (r) 9x2+u Factor 16a10-256b8. Let us consider the factor 4a5-l6b“. Notice that there is a common factor here. Or that both of these terms may be divided by the same number. The common factor here is Then ue5elob4 can be written as 4(a5-16bu). Is there a common factor in 4a5+16bu? If so. what is it? Then (ue5+16b“)(ue5-16b“) e 4(a5+4b4)4(a5-4b4). Putting the numerical factors together. we get 16(a5+4b4)(a5-4b4). - 117 - 66. ones where there is a 67. common factor and polynomials which are the difference of two squares. 67. Removing the common factor 68. first we get 4(4p4-q6). Factoring 4pL"-q6 as the difference of two squares, Isn’t-m6 = 2+<2p2+q3><2p2-q3>. This could have been factored as the difference of two squares getting (%2-2q3)(4p2+2q3). Each of these factors has a common factor and when we remove this We get the same result as above. 68. 3(p2-25) = 3(p-5)(p+5) 69. We shall factor polynomials completely which means that none of the polynomial factors will be factorable by any of the methods that we know. So far. we know how to factor what kinds of polynomials? In 16a10-256b8. we could have removed the common factor first. Both of these terms are divisible by 16. So 16e1°-256b8 = 16(alO-16b8). Then alO-léb8 can be factored as the difference of two squares. We would obtain exactly the same set of factors although the factors might be arranged in a different order. Factor lépuu4q6 . In some cases we must remove the common factor first. Consider 3p2-75. These terms are ngt,squares, so we must see if we can factor by any other means. Notice that there is a common factor. Remove the common factor and then factor again if possible. Repeat this process until none of the factors can be factored again. 3p2-75 = = Factor completely. (a) su-topzqz (b) 4a6-24a3 (e) 4x2+y2 (d) 3x5y-483W3 (e) 189m8n3-105m6n _ 118 - (r) 45aub5-63a3b7c4-9a2b3 (g) 4y(p-q)+2(p-q)-X(p-q) (h) 2x3-8x- 3x2y+12y (i) x3-2x2-4x+8 69. (a) (8-7p%)(8+7pq) 70. We obtained the factors (b) 4a3 (xe2)(x+2)(xe2) as the result (c) This can't be factored. to part i of the last frame. (d)3xy 3mg 2?“ 4W) (e) 21m n(9mg -5) Can this be written as (x#2)(xh2)2? (r) 9a2b3(5a2b2 7ab4o+l) Is (xn2)(xs2)2 in factored form? (a) (p-q)(4y+2-X) (h) 2x(x2-4);gy(x2-4)= (2x-3y)( (Zthy)(XP2)(X#2) (i) x2] Factor lému-(5+3n)2. '7E3.. [4m2+(5+3n)][4m2-(5+3n)] 77. When parentheses and brackets are 77 . [4m2+5+3n] [41112- 5-3n] 78 . 78. ( 5b“+[5a2-10])(5b4-[5a2-103)= 79. (5b4+ 55.2-10 ) ( 5b4.5e2+1o) 79. Both of these factors can 80. be factored again. Both have a common factor 0f 5. used. it is sometimes necessary to remove whichever set is innermost. write the answer to the last frame using only the brackets. When we factored 16m4-(5+3n)2. we wrote the sum and difference of the square roots. Note that when we did this. we considered the square root of (5+3n)2 as (5+3n) 22.1; sen. Whenever the polynomial consists of more than one term and it is to be connected to another number by one of the operations addition, subtraction, multiplication or division, we shall treat this polynomial as one number and enclose it in parentheses. If the result than contains more than one symbol of grouping, we generally write the result so it contains at most one symbol of grouping. Find the factors of 25b8-( 5.12.10)2 and write the result so it contains at most one symbol of grouping. Look at the factors in the answer to the last frame. Can either or both of these be factored again? If further factoring can be done. how would .you proceed? Factor 25b8-(5a2-10)2 completely. - 120 - 80. (5b4+5a2-lo)(5b4.5e2+10) = 5(b“+a2—2)5(h4-e2+2) = 25(b“4a2-2)(bu-a2+2) 81. (a) (x2-3> (b) (2p-5)-6 (2p-5+6) = (2p-5-6)(2p-5+6) = (ZP-ll)(2p+l) (12+3p+6)(12-3p+6) (12+3p4-6 ) (12-3p-6 ) (18+3p)(6-3p) There is a common factor of 3 in both factors. (18+3p)(6-3p)=9(6+p)(2-p) (x~3)+(y+4) (x~3)(y+4) = xs3+y+4 xs3-y-4 = (x4y-v-1)(x-y-7) (2yr7)+(3y+8) (2y+7)-(3y*8)= 2y47+3y+8 2y+7~3Y-8 = (5y+15)(-yrl) The factor of 5y+15 has a common factor of 5. (5y+15){-y-l)=5(y*3)(-y-l) This can't be factored as it isn't the difference of two squares. This isn't the difference of two squares. but it can be factored as the two terms (0) (d) (e) (r) (g) 82. 81. When there are several numerical factors, we usually multiply them so that the final result has only one numerical factor. This is not absolutely necessary. however. Factor the following completely. Use only one symbol of grouping in the final answer. (a) x5-8l / (b) (2p-5)2-36 (c) 1nu-(3p+6)2 (d) (x-3)2-(y44)2 (e) <2y+7>2-(3y+8)2 (f) (3p-1)2+36 (a) 4x3(y2-l)-9(y2-1) (h) p6-9p”-16p2+144 (i) x9-25 x3-8 is not the difference of two squares. x3-8 is the difference of two cubes. In order to determine whether a number is a cube, we must first have the definition of a cube root. The cube root of N (written Q/fi) equals a only if a‘a'a = N What is the cube root of 8? Why? because . have a common factor of (yZ-l). 5%2. 83- 2°2-2oz-2-2 - 121 - (y2'1)(l+12-9)0 Both of these factors can be factored as the difference of squares. Don't forget that the square root of l is l. (yZ-lxurze) = (y+l)(y-l)(2x-3)(2x+3) (h) p 1:ng 2-9)- 416( 2-9) = )(p4-16E = (p-3)(p+3)(p +4)(p-2)(p+2) (i) This can't be factored. x is not a square. Don't forget you are dealing with x9 m 9. 2 because 2°2°2 x3 x because x‘x‘x 84. 8 83. We shall use much the same method to find cube roots as we did to find square roots. First factor the number and then use the definition. To find the cube root of 64. first factor the number. You might say 64 = If you recognize that 8 is a cube, you can take the cube root of 64. ‘Q/EE =\§/§7§ = 2'2 = 4 You might not recognize 8 as being a cube. In that case continue factoring until you have numbers whose only integer factors are the number and 1. Express 64 as the product of integers whose only integer factors are itself and 1. Since by definition. QJN = if a'a°a = a only N. 2'2'2 must be a cube. W(2'2-2)(2'2-2) = 2-2 = 4. Numbers whose only integer factors are itself and l are called prime numbegs. In frame 84 then we expressed 64 as the product of prime numbers. O 2 3 .703 2 7 AAA vvv N \i) \7 a b c '2 The order of the factors is immaterial. 85. factor 216 86. 2.3.2.3.2.3 or 8°27 01‘ 8030303 8?. V2'2'2'3'y3 = 2'3 = 6 {/8'2‘7 == 2'3 = 6 88. ‘3/0'3‘3527 = 3'3 = 9 89. W<5°5°5><3°y3> = 5'3 -= 15 9o. 90. (a)W3°3°3'8'8 = 3'2'2 = 12 91. (b)\Z/2'2'2'2'2'2°8'8 = 2g2°2'2 = l - 122 - 85. 86. 87. Empress each of the following as the product of prime numbers. (a) 42 (b) 63 (e) 56 To find the WEI, first ' . The question might arise as to how we factor this number. Certainly 216 = 2'108, and 2 and 108 are factors of 216. However, we must consider why we wished to factor 216. Our original problem was to find the cube root of 216. Whenever we are to find the cube root of a number. we factor it until we have prime factors or until one or more of the factors is a number you recognize as a cube. Factor 216 so that we may then find the cube root of it. Now find the cube root of your set of factors. 88. QU§E§'= = 89.‘QJ§§§3’= = (a)\Q/57§§'= (b) Q/409g = (CM/173 = If you are going to do much algebra. you would find it useful to know the squares of the numbers 1 through 25 and the cubes of the numbers 1 through 7. “A"- _ 123 - (c) Remember this is a square root andVN = a only if a'a = N _—_—J_——. Vaayrw=2ev=42 91. cubes 92. 92- x-2 93. 93. divide x3-8 by x-Z. 91., Remember if a°b = c. then 2.: b a O 94. The other factor is x2+2x+4. 95. x3-8 = (xr2)(x2+2xw4) 95. difference of two cubes 96. 96. 2p2 because 2p2(2p2)(2p2)=8p6 9?. 3q3 because 3q3(3q3)(3q3)=27q9 97, 2p2-3q3 because one factor 98. of the difference of two cubes is the difference of the cube roots. T3 factor a polynomial such as x -8. first determine whether this is the difference of two squares or the difference of two cubes. That is. find V3.5 and the VE’ if you can. If both of these do not have square roots. then find Q/x3 and W8 if you can. In this case, x3 and 8 are both (squares. cubes) choose one One factor of the difference of two cubes is the difference of the cube roots. Therefore. one factor of x3-8 is Since x3-8 has a factor of x92 or x3-8 = (x~2)( ). how would you find the other factor? Find the Sther factor and then express x -8 as a product. Does 8p6-27q9 represent the difference of two squares or the difference of two cubes? Find QJBpé and 27g? and tell why they are the correct cube roots. One factor of 8p6-27q9 is because one factor of the difference of two cubes is (See frame 93 if necessary.). If one factor of 8p6~27q9 is 2P2-3q3o the other factor is Express 8p6-27q9 as a product. -12u- 98. other factor is 4pu¥6p2q3+9p6. 99. Express 216- p3q 12 as a product. 100. 101. sp6-27q =<2p2-3q3>(4p”+6p2q3+9p6> This is the difference of two 100. cubes. cube root of 216 is 6. cube root of p3q12 is pq“. 216-93912=(6-pq4)(36+6pqufp2q8) 6ny3-17=(by.3)(loy2+12y+9) The binomial factor of 4yb3 is the difference of the cube roots. In the trinomial. 16y2 is the square of 4y; +12y is the product of 4y and -3 with the opposite sign; and +9 is the square of -3. Be careful here. y6 has a cube root of y2 but 25 doesn't have a cube root. NOW examine them to see if they both have square roots. V53=y3 andV'2—5= y6-25 = (y3f5)(y3-5) 101. 102. The second factor in these examples is a polynomial which has three terms. A polynomial which has three terms is called a trinomial. In (6-pq“ >(36+6pq“ +p2 q8). notice that the first term of the tri- nomial. 36, is the square of the first term of the binomial. 6. The second term of the trinomial. +6pqu, is the product of the two terms in the binomial. 6 and ~pq . with the opposite sign. The éhird term of the trinomial. q . is the square of tfie last term of the binomial. -pq . This is true when we factor the difference of two cubes. Factor 64y3-27. y6-25 = when expressed as a product. If you have forgotten how to factor the difference of two squares. go back to frame 61. hthis %ast frame. we had a number y which was both a perfect square and a perfect cube. You must examine both terms before proceeding to factor. If both terms are not cubes or if both terms are not squares. then you cannot factor unless there is a common factor. (a) 8p6-64q9 (b) 4m6-64n12 (c) 8x6—49 Factor: - 125 - 102. (a) This is the difference 103. In 41116-64m12. we factored this as of two cubes. 8 and q9 the difference of two squares and are cubes. p6 and 64 are then found a common factor in both cubes and squares. All_the numbers must be squares or all the numbers must be cube. (2p2-4q3xup Lap 2013.16.16) Notice that both of these factors have a common factor. 2 is the common factor of the binomial and 4 is the common factor of the trinomial. Removing these. we 8(p2-2q3)(pg+2p2q3+4q5) This is the difference of t o s ares. 4 is a square. mg. 64 and n12 are both squares and cubes. All the numbers must meet the same condition. (13} each of the factors we had. we could have removed the common factor first, and 4p6-64m12 = 4(p6-16n12). Then p6-l6n12 can be factored as the difference of two and equals 4m6-64n12 = (2m3-6n6)(2m3+6n6>. Removing a common factor of 2 from6 each 3fagtor we get (m3-4n6 )(m3 +4n This can't b factored. 8 is a cube. is both a cube and a square. and 49 is a square. There is no common factor. (c) 103. Squares 104. P6-l6n12 = (p 3#4116 ) (P3 +4n6 ) 10L“ 3(X-3)(x2+3x+9) 105. We would ldet the same factors for M—64n as we had in the answer to frame 103 part b. It makes no difference in which way we do the factoring, however if there is a common factor. it is usually to your advantage to remove it first as then you have smaller numbers to deal with. Factor 3x3-81. The sum of two cubes can be factored into a binomial and a trinomial. The binomial factor is the sum of the cube roots of each of the temps. 105. 2m+3 106 . 4m2 found by squaring the first term of the binomial factor or by squaring 2m. lO?.-&n -126- 106. 107 . 108. found by finding the product of the two terms in the binomial and using the opposite sign or by multiplying 2m by +3 and using a - sign. +9 found by squaring the last term of the binomial factor or by squaring +3. (2mi3)(4m2-6m+9) 108. 109 . 110. (6+5x) ( 36-3Ox+25x2) 111. difference of two squares flflmn of two cubes difference of two cubes This is the sum of two cubes. 109 . 110. 111. The trinomial factor is formed in exactly the same way as it is formed when we factor the dif- ference of two cubes. (Refer to frame 100 if necessary.) For example. to factor 8m3+27. we first check to make sure that all the terms are cubes. Then we can form the two factors. The binomial factor for 8m3+27 is . The first term of the trinomial factor is and is found by . The second term of the trinomial factor is __ and is found by . The last term of the trinomial factor is and is found by . Thus. 81113427 = when expressed as a product. Factor 216+125x3. We have had four kinds of polya nomials which we may factor thus far. They are , The sum of two squares is pg; factorable. 'we will discuss this in more detail later. - 127 - and polynomials which have a common factor. Remember this last category includes polynomials which don't have a common factor in the original but which do have a common factor when terms are grouped together. 112. (a) (62 3xy2)(36+18 +gn2y y“): 113. 27 7(2-xy 2x§4+2 +xy (b) 4 x3—4)( (c)x (x2-16)+l(x2-16)= (x2-l6)(x3+l)= (x44)(x-4)(x+1)(x2-xwl) (d) cantF 't be factow (9) (5a n—b3c2)(25 5a §y2a4b3cz+b6c c”) (f) (3+2y2)(9-6y2+u (a) (4x2 -l)(y3 +8) (2x»1)(2xel)( +2)(fi (h) y3(y2-4)-64(y = +8(t+x2-1> (h) y5-uy3-6ay2+256 (i) y5-4y3+64y2-256 (j) m6+l6 (2Xh3)3-y3 is the difference of two cubes and so can be factored. Q/(Zx-B)3 because because +4) -u> = (y ~4)(y3-64) Factor (2x-3)3-y3 Express this result using only one sign of grouping. Factor 8y3+(3y-5)6. Express this result using only one symbol of grouping 117. 118 o 119. 120. 121. - 128 - Emmy-sfltuyZ-zycysWay-5)“) 118. Factor the following. [2y+9y2-30y+25][uyZ-2y (9y 2-30y+25)+81y4-540y3+1350y2 -l500y2-1500y+62 (9y2-28y+25)(8 ~558y3+lfilky2 ~1550y+6 2 5) (a) [(2y+5)+(y+l)l [(23146)2 -(2y*5)(y+1)+(y+1)23 -[3y#6][EyZ+29y+31]= 3(y+2)(7y 29y+31) Don't forget to remove common factors if there are any or to refactor by any of the other methods we've had if they apply. (b) [133-2 -J[1oy6+ay3+ 32 [uyB-uy‘r-rfluoyé +16y ".12y3-1-16y2 -2 hyr9 ) [(2 3-5)- (4 3+ )1 (c [(chg-E)§+()2Cx3§5>(4x3+3)+ 4 E-2x3-33[28x6-10x3+l9]= 2(-x3-u)(28x6-10x3+l9) 3 terms because it has 3 terms 2Y*5 and y+l (Either order) 120. 121. 122. Express your results using only one symbol of grouping. (a) (2y+5)3+(y+l)3 (b) suy9-3-(ux3+3>3 H9. In this last frame you multiplied several binomials together. These binomials had similar terms. For example. (2y*5) is an instance of multiplying two binomials which are similar. 2y+5 and y2+2 are ngt’similar. They are of different degrees. 3a-b and 2a+3c are not,similar. Even though they are of the same degree. the terms in 3a-b are ngt’similar to those in 2a+3c. When you multiply two binomials which are similar such as 2y¢5 and y+l, how many terms are there in the product? In (2y+5)(y+l). the product is 2y2+5y+2y+5 which equals 2y 2+7y+5. This is a trinomial. Why? Since (2y*5)( +1) = 2y2+?y*5. the factors of 2y 7y+5 are and . In order for a trinomial to be factorable. it must be of Quadratic szm. To determine whether a trinomial is of quadratic form. arrange it in either ascending or descending order and put.in any missing terms. Fbr example. l+x6~x3 could be written x6+0°x5+0'x4-x3+0°x2+0°x#l. - 129 - 123. If the terms are arranged in descending order. the last term will be a number without a variable. If the terms are arranged in ascending order: the first term will be a number without a variable. If the polynomial is of quadratic form, there will be the same number of missing terms between each successive pair of given terms. In the above case. there ar two missing terms between x8 and x3 and there are also two missing terms between x3 and 1. so this polynomial is of quad- ratic form. Which of the following are of quadratic form? (a) x9+4-3x (b) x4+2x2—3 (c) x3-2-3xi (d) Xh3x2+2 (e) 2.8-6-79 Not all trinomials of quadratic form are factorable. If a trinomial is of quadratic fOrm and is factorable. it will factor into two binomials. Fbr*example, x+3x2+2 is of quadratic form and equals (3x»2)(-x»1) because (3x+2)(-x#l) has a product of -3x2+x#2. In multiplying two binomials, we follow certain steps. 'We shall review these as we will reverse them in order to factor the trinomials. List the steps used to get each term of the product (3xe4)(5x»2). - 130 - 123. the product is 15X2-14Xh8. 124. Then to fac r xz-Bxfiz. we will factor the to get the first (a) multiply the first terms term of each binomial. of each binomial to get 15x2 (b) find the two cross products x2-3x+2 = (x )(x ). and add them to get ~14x (c) multiply the last terms of we can't use the second term of each binomial to get -8. the trinomial, ~3x, at this time as it is the result of adding the Review frames #6 and #7 if two cross products and in order necessary. to find the cross products we need both terms of each binomial. The last term of each binomial comes from what term in the trinomial? 12h. From the last term of the 125. When we factor the last term of trinomial which is +2 in a trinomial. we will consider this case. only the absolute value of the number and then put in the algebraic signs later. Thus. in this case. we shall factor 2. The only possible factors of 2 are 2 and 1, so now we have xz-waz = (x 2)(x. 1). Now consider the cross products. In this case the cross products are and . 125. X and 2x (Either order) 126. Our problem now is to put in the correct signs so that the cross products equal the second term of the trinomial which is ~3x in this case. These signs must also be such that the product of the last terms in each binomial equals the last term of the trinomial or +2 in this case. Supply the necessary signs so that x + 2x = --3x. 126. -x 4. .2x = -3x 127. Our problem then has become x?-3x#2 = (xe2)(xrl). Now we must check to see if the product of the last terms in each -131- binomial equals the last term in the trinomial. Does -2 multiplied by -l eqyal +2? Yes, it does. Thus (KHZ) and (xel) are the correct factors. To factor x2-ux.12. we first determine that it is a trinomial in quadratic form and then we know that if it is factorable, it will factor into two binomials. Once we have determined this, we will and then 127..factor 2x and then factor 12. 128. Put in the first terms of each binomial. radix-12 = (_ )(__ ) 128. (x )(x ) 129. Now we must determine the factors '% of 12. What are the factors of 12? 129. There are several pairs of 130. Our problem now is to choose the factors of 12. correct pair of factors of 12. 6 and 2 3 and 4 We have a choice of: l and 12 (x l)(x 12) (x 2)(x 6) (x 3)(x 4) The second term of the trinomial which is -hx in this case is the result of adding the cross pro- ducts of the binomials. Thus we must find the cross products for our 3 possibilities. For (x l)(x 12), these products are . For (x 2)(x 6). these products are . For (x 3)(x. 4). these products are . 130. 131. - 132 - x and 12x 2x and 6x 4x and 3x The numbers must have different signs or one number’must be positive and the other must be 132. 131. We must now consider the signs we are going to use. In the given trinomial, x2-uxe12. the last term is ~12. In order for the product of the last terms in each binomial to be a negative number, what must be true of the numbers which are multiplied? For (x 1)(x 12). we must consider (x—l)(x&l2) and also (xwl(xe12) as both -l(+12) and l(-12) have a product of -12. Our trinomial was XZ-uXhlZ. NOW‘We only need to consider whether or not we can get the -4x as we chose the first terms in each of the binomial factors so that we could get x2 and we also chose the last terms in each of the binomial factors so that their product was -12. negative Do either (x-l)(x+12) or (Ml) (x912) e ual the given polyh nomial ~4Xh12? Since neither (xel)(x»12) nor (x*1)(xelZ) :56 the correct factors for -hxe12. what should we do now? 132. No. The sum of the cross products in the first case is +llx and in the second case is -1lx. 133. Find the correct factors of XZ-uXFl2. Consider (x 2)(x 6) to see if we can.put in the correct signs to get our given poly- nomial. If neither possibility gives us the required result. 'we will then consider (x. 4)(x 3). If neither possibility works here. we can say that x2-hxelZ is not factorable or can't be factored. 133. 134. 13%. (xfi2)(x>6) 135. List all of the possible sets of factors that there are for 22.-emu. 135. First factor x? and then 136. To decide which is the correct factor 24. set of factors. we must Remember that the last term of the trinomial is negative, . so we need numbers with different signs. 136. 137. 138. 139. 140. -133... Possible factors: (xel)(x*24) (x#1)(xe2h) (xe2)(x512) (x*2)(Xh12) (X-BXW Ex+3)(x-8 x-£+)(X+6) (ml-LPXX-é) decide which pair of factors 13?. gives ~6x which is the second term of the given polynomial. None of the possibilities 138. listed in frame 135 give x2-6xp24 and since these are the only possibilities. x?-6xe2# cannot be factored. Possible factors are 139. (xr1)(xw16) (x!l)(xel6) (x#2)(xe8) (x-2)(x+8) (x—h)(x#4) Correct set of factors is (xe8)(x+2) Arrange it in order and check 1&0. to see if this is of the quadratic form. xgellx2+18 is of quadratic 141. form since there is one missing term between each pair of successive terms when the polynomial is arranged in order. Note that since the last term is positive. the last terms in each of the binomials must have the same sign. Possible factors: (x2-l)(x3-l8) (x2+l)(x2+18) (x?-2)(x2-9) (x2+2)(x2+9) (x2-3)(x2-6) (x2+3)(x2+6) Correct factors are (x2-2)(x2-9). Factor x2-6xe2fi. Sometimes the factoring of a trinomial of the quadratic fdnn is referred to as the trial and error method. It is obvious why this is so. Factor Xg-6Xh16. To factor x9+l8~11x2. what do we need to do before starting to form the possible binomial factors? Factor x4+18-llx2. In frame 66. we stated that when we factored, we usually expressed the result in terms of factors which could not be factored again - at least by any of the methods we have had. Look at our last result. nth-1123448 = (x2-2)(x2-9). Can either or both of these factors be factored again and if so, which one(s) and why did you decide it could be factored again? -134- 1&1. x2-9 can be factored again 1&2. Complete: because it is the difference & of two squares. x -llx2+18 = (x2-2)(x2-9) = x2-2 can't be factored again. 1&2. (x3-2)(xs3)(x»3) 143. Factor 12+xex2. Do this twice - once when the terms are arranged in descending order and once when the terms are arranged in ascending order. 1&3. In ascending order we have 1&&. Since we factored the same poly; 12+xhxg = (&-x)(3+x) nomial in each case. all our results must be equal. Arranged in descending order, we have that 12+x+x2 = That is. (&-x)(3+x) = (-xw&)(x+3) x?+x#12. = (xa&)(-x+3). -x?+x§12 = (-x+&)(x+3) Consider (&-x)(}fix) and or (-x»&)(x+3). It is quite - 2+mwlz = (x»&)(-x-3) obvious that these are equal since &-x equals -x#& as these are exactiy the same except for the order in which they are written and certainly 3+x and n+3 are the same. Now consider (-x»&)(x#3) and (x-&)(-x~3). The factors -x#& and x9& are n21 equal and.x#3 and -xe3 are g2; equal. Even though the individual factors are not equal, we have said that (-x*&)(x+3) and (Xh&)(-Xh3) or that their products are equal. How would you determine if this is a true statement? That is. does (-x#&)(x+3) equal (x-&)(-x-3)? lhi. Find the products or multiply 1&5. (-x»4)(xe3) = and the factors to see if we obtain the same polynomial (xe&)(-xs3) = . in each case. Are these the same polynomial? 1&5, -x2+m»12 is obtained in both 1&6. There sometimes are several cases and it is the same different sets of factors for a P°lynomial. given polynomial. If you obtain a different set than the ones given in the answer, how can - 135 - you check to see if yours are correct? 1&6. Multiply them together to 1&7. There is another technique that see if you get the original we may use. polynomial. ‘We had (-x+u)(xe3) and (xe&)(-xp3). Consider the first factor in each case. In -x&&) and (x>&) what similarities and differences do you note? 1&7. The numbers x and & are the 1&8. If we factored (-x%&), what same but each one has the could we use as a common factor opposite sign so that we would get (?)(x~&)? or all of the signs in one factor are the Opposite of the signs in the other factor for the numbers x and &. 1&8. use a factor of -1 because 1&9. Consider the second factor in each of (-x&&)(x+3) and -l(xr&) = (-X+&) (xr&)(-X~3). Factor (xw3) so that it equals some number multiplied by (~xs3). 1&9. (xw3) = ~1(~Xh3) 150. Then (-x#&)(x#3) can be expressed as -1(Xr4)(-1)(-Xs3). By rearranging the order of these factors. we have -l(-l)(xp&)(-x+3) and if we multiply the numerical factors we get +1(xs&)(-xs3) or (xs&)(-x-3). In frames 31 and 33 of Chapter 2. we defined the additive inverse or the negative of a number. To review, the additive inverse or the negative of a number is a number such that the sum of it and the original number is 0. Could we say that (-x#&) and (xe&) are the negatives of each other? Why? - 136 - 150. yes because their sum is O. 151. The product of twg,numbers is always equal to the product of the negatives of these two numbers. Let us consider the numbers -2 and 3. What are the negatives of these numbers? 151. 2 and ~3 152. According to our statement in frame 152, the product of our two original numbers. -2 and 3, should equal the product of their nega— tives which are 2 and -3. Does the product of -2 and 3 equal the product of 2 and -3? 152. yes. 153. Consider the polynomials (x92) and (x2+3). The negative of (xsz) is The negative of (xa+3) is 153. (-x&2) or (2-x) l5&. Multiply and see if the product of the two original polynomials (-x2-3) equals the product of their negatives. 154. Yes 155. Use the statement in frame 152 product is x3-2x2+3x~6 in to write an equivalent set of both cases factors for (~x+3)(xe2). 155. (xe3)(-x#2) 156. To return to our factoring problem of 12+x-x2. (xe3) is the negative of (-x+3) (-xw2) is the negative of (xe2) ‘We can say that (&-x)(3+x) equals (Xh&)(-Xh3) because the product of two numbers equals the product of the negatives of these two numbers. 'Warning - YOu must have the negatives of both.numbers. 'We have only talked of Eng factors. Is (-3+X)(3+x) equal to (B-X)(3-x)? Give a reason for your answer. 156. NO. 157. Find the products and verify the (3—x) is £93 the negative of answer to the last frame. (34-20. -137... 157. (-3+x)(3+x) = ~9+x2 158. (3-x)(3-X) = 9-6x+x2 158. NO. 159. (x-Z) is p_o_t the negative of (Zix). (3-x)(2+x) = 6+x-x2 (x~3)(x~2) = XZ-wa6 159. (a) (xr3)(x+2) 160. (b ) (8-x)(&-x) or (xe8)(x~&) (c) Afliggge in order first. .6 = ( fl+6)(x2-l). Factor again to get (x2+6)(x-1)(x+l)g (d) Arran e in order first. (e Ef (E (i ) ) ) ) ) -96 = (x3-12)(x3+8) x3+8 is the sum of two cubes. Factoring a ain, we et (x3-12)(x+2%(x2-2x4&§ (xwl6)(xw2) (x&12)(x+8) can't be factored (xelO)(xs8) :firag @e in order. -21x3+80 = (x2-16)(x2-5) x2-16 is the difference of two squares. Final result: (xe&)(x»&)(x2-5) (-3+XD(3+X) = (B-X)(3-X) = Does (3-x)(2+x) equal (xs3)(xs2)? Give a reason fer your answer. Find the products and verify your answer. Factor completely. (Refer to frames 12& - 1&3 if necessary.) (a) xz-Xsé (b) 32-12x+x2 (c) xu-6+5x2 (d) ~96+x6-&x3 (e) x2+18x§32 (f) x2+96+20x (a) x2-6xm7 (h) 28-18wa (1) 8o+x4-2lx2 Suppose you were to find the factors of 2x.-l3x»15. This is a trinomial of quadratic form. so you should try to factor it. If it is factorable, you will obtain two binomial factors. The first term in each binomial must satisfy what condition? - 138 - X4-21x2480 = (x2-16)(x2-5) x2-16 is the difference of two squares. final result: (x-&)(x+&)(x2-5) 160. Their product must equal the 161. Fill in the first terms in each first term of the trinomial binomial. of 2x2 in this problem. 2x2-13x+15 = (_ >(___ ) 161. (2x. )(x ) 162. Now we must determine what numbers will be the last terms in each binomial. Since the last term in the tri- nomial is +15. we can use +1 and +15, -1 and —15 +3 and +5 or -3 and -5. Remember since we have a +15, both of the factors must have the same sign. The first terms in the binomials have different coefficients. Therefore, if we use the factors of +1 and +15. we must try both (2x#1)(x#15) and (2x+l5)(x+l). You can see that the sum of the cross products is different in each of these. Are either (2x+1)(x+15) or (2x#15)(x§1) the correct set of factors for 2x2-l3x415? 162. No. 163. What other possibilities are there? Which one, if any, is the required one? 163. Other possibilities are: 16&. Factor 6x2+5xp&. (23>1)(x-15) (2x-15)(x-l) (2x95)(x-3) (2xr3)(X~5) (2x*5)(xm3) (2x#3)(X*5) correct set is (2x-3)(x-5) [ere are many possibilities are as 6x2 has two sets of Lctors and -& has two sets ' factors. fi+5xp& = (3x+&)(2x-1) the second one :h the 3x - 139 - 165. 166. There is one fact that will make less of a trial and error process when factoring a trinomial. If the trinomial has a common factor. remove it first. Then the resulting trinomial will have no common factor and hence its factors will not have a common factor. For example, when factoring 6x2+5xe&, sup ose we use as factors of 6x . 2x and 3x. So far, we have 6x2+5x-&= (2X )(3X ). Now we will use the factors of -& in order to complete the binomials. Let us consider -2 and +2 as the factors of -&. 'we cannot use the -2 with the 2x. having (2x—2), as then this factor has a common factor and our given polynomial of 6x2+5x~&, doesn't have a common factor. Likewise. we cannot use the +2 with the 2x getting (2xfi2) for the same reason. Then -2 and +2 are not the factors of -& when we consider 2x and 3x as the factors of 6x2. Keeping the factors of 6x? the same - namely 2x and 3x. we will tiy -l and +& as the factors of If -1 and +& are the correct ones, in which of the binomials. (2x )(3x ), must the +& be put? It can't go with the 2x giving a factor of 2x#& as this now has a common factor and this can't be as long as the original poly- nomial doesn't have a common faCtoro Factor 2x2-xy-6y2. ix+3y)(x-2y) ) (3p+5)(2p-l) ) (7m-2)(9m-l) > 2xy2(4xs3)(3x~4) > (x2-9> = (maxx-sxx-zxxm > (81x3+l)(x3+l) s (81X3+1>(X+l)(k§ -x+1) D (&m2-5n)(7m2+2n) ' (8xr3)(3x%&) . Eng -(7p—4)][2p2+<7 ~4)]= (2p2 2-7p+u>( 2p2 2+7p-43= (2p2 -7p+#9(2p-1)(p+40 (xr3)(x2-[&x-31)= (xr3)( ~4xs3) = (xr3)(xr3)(xsl) tb)-&][(a+b)+21 a+b) and +2(a+b) -140- 167. 168. 169. 170. Factor completely. (a) 6p2+7p-5 (b) 63m2+2-25m (c) 24x3y2—50x2y2+24xy2 (d) x9-13x2+36 (e) 81x6-l-80x3 (f) 28m4-27m2nelOn2 (g) 2&x2+23X-12 (h) apis<7p-u>2 (i) x2(X-3>-(X~3)(&x-3) (a+b)2-2(a+b)-8 is a trinomial in (a+b) if we do not remove the parentheses. we can factor it as we do any other trinomial. First terms in each binomial are the factors of (a+b)2. or (aib) and (a4b). (a+b)2-2(a+b)-8=[(a+b) ][(a+b) ] Next we put in factors of 8 remembering to use numbers with different signs as we want the product to be -8. Last we examine the cross products so that we obtain the correct second term of the trinomial or -2(a4b) here. Finish factoring. Let us consider the cross products here. These are __ and . These are similar terms since the factors other than the numerical ones are the same. SO'We can add by adding the coefficients and multiplying the result by the - 1&1 - common factor, getting -2(a+b). ‘Write these factors using only one symbol of grouping. 170. (a+b-u)(a+b+2) 171. Factor 2(xay)2-3(xay)+l. 171. [2(xpy)-1][(xéy)-l] = 172. It is usually a good idea to keep several signs of grouping in the [2x—2y-l][th-1] first step of a problem where you have a group of terms used as a single number in the original. If it is necessary to have only one symbol of grouping in your final answer. then remove the other symbols of grouping in succeeding steps. Ybu will probably make fewer errors this way. Factor &-3(2x#y)-(2x#y)2. Use only one symbol of grouping in your final answer. 172. [&+(2xwy)1[1-(2x«y)] : 173. Factor completely using only one symbol of grouping in your final [&+2x#y][1-2x~y] answer. (a) 6(x—y>2-5(x-y>-6 (b) 3(2a+b)2+8b(2a+h)-3b2 (c) 15(x-2y)2+1&(x-2y)-8 173. (a) [3(xsy)+2][2(xry)-3] = 17&. Factor completely. These [3xr3y+2][2xp2y-3] involve all of the different kinds of factoring that we have (b) [3(2a+b)-b][(2a+b)+3b] 2 had. [6a+3b-b][2a+b+3b] = (6a+2b)(2a+&b) = (a) 6a2b2-66a3b‘+-72a%6 &(3a4b)(a+2b) (b) 6xu-7x2-l (e) [5(xr2y9-ZJEBCXP2y)+&1 = <5x-10y-2><3x-6y+u> (c) user (d) 27a3-75ab6c4 (e) 12-23(p*q)+10(p+q)2 (r) 3xl7+9x9-12x (g) (2x?-8t2)(2-5x)+3x2(2x3-8t2) (h) x7-8x9-4x3+32 - 1&2 - (i) l6-(2xa3y)2 (j) 64-(3p+q)3 (k) x3+16 (l) 4m2+9 (m) 10.3(r-s)-27(r-s)2 (n) 2x6-8xu-x244x 174. (a) 6a2b2(1-12ab2)(1+ab2) (b) can't be factored (c) (5+4y2)(25-20y2+16y“) (d) 3a<3a-sb3c2>(3a+5b3c2> (e) [3-2(p+q)1[&-5(p+q)1 = (3-2p-2q> (4-5p-5q) r) < 8-1>< 8+4) = ( §fc(:u+l)(:2+l)(x-l)(x+1)(x8+&) (s) (2x3-8t2)(2-5x»3x2) = 2(xr2t)(x+2t)(2+x)(l-3x) (h) xi x3-8).u(x3-8) = ( -8)(x#—u) a (x-2)(x2+2x+&)(x2+2)(x2-2) (i) [4-(2x-3y)][4+(2xa3y)] a (4-2x+3y)(&+2xe3y) (') [LP-(3p!- )1[16+&(3p+ )+(3p+q)23 = J (4-3p-qc)l(l6+12p+4q+gp2+6pq+q2) (k) can't be factored (1) can't be factored (m) [5-9(r-s)][2+3a—2= 3.92:3). 3a (0) 2 = m+3n (man) (m—n) = SZa—b}§2a+2b2 (d) 2a-b ZaZa+b5 (e) néxpfizgéx+§zg = x x+2y Zx-y X( 2x-y7 How would you check to see if 2 : 2m—n 7 m§3n 1m+3n men‘ to see if Qéx-jzxgéx-IQX; = Mfr-12;? x x+2y 2xey x.2xey -149- 21. If 2(m+3n)(m~n) = (m+3n)(m_n)2 22. In other words, we may divide both then these are equivalent frac- tions 0 Remember,‘g =‘g b d b and d don't equal 0. Refer to frame 122, Chapter I if necessary. if ad = be and If “(x-3y)(x+2y)X(ZX-y) = x(x+2y)(2xey)4(xe3y) then these are equivalent fractions. Note that in the above cases, the factors are the same, but occur in different order. The commutative and associative laws of multiplication allow us to say that the two results are equal. 22. (a) Divide both numerator and denominator by 3x getting x 2x+ 2 3+2y (b) divide both n erator and denominator by 6a b getting 1,2 (0 divide both numerator and denominator by 3q(p+3q) getting 2 2 - 2p+5q 23. are equal if ad = bc, see if ad = bc when 16 is considered -21 as a/b and.:l§ is considered as 21 as C/de 24,{_2’ because 3(2) = ~2(-3) or -2 because the numerators and and denominators of the two fractions are the negatives of 25. -4 = a nwmerator and denominator by a common factor or we may multiply both numerator and denominator by a common factor to get equiv- alent fractions. If the numerator and denominator have ng'common factors, then we say the fraction is in its lowest termS. Reduce each of the following to lowest terms. (a) 2x2§2x+§§2 x 3+2y (b) - 42e2b3 Zhaub (c) 6292522:§§%92:29; - 3a p+3q 2P+5q In the answers to parts b and c, if you have the "-" sign in a different place, your answer may be correct. Check it to see if ad 2 bc when the fractions are considered as a/b and old. Let us consider another useful technique in determining equiv- alent fractions. How would you decide if :lé 21 is equal to‘ilé ? -21 Since two fractions a/b and c/d 2n. -l6(-21) equals 16(21) so -21 21 ' Any two fractions where the num. erators and denominators are the negatives of each other will be equivalent fractions. (Review frames 31 - 37 of Chapter 11 if you don't remember what the negative of a number is.) :2:... 2 _2 because . because -3 7 3 7 (J?) (5X3) (7) = (-3)(7)(‘*) (5) . You will notice that (4)(5) is 25. the negative is (3)(-4)(-2) 26. 27. yes. each other. OR is (-3)(u)(-2) on is (-3)(-4)(2). i; °t-§i-E%y - 15o _ WOrk out the fractions. All have a value of -6/35. 26. 27. 28. 29. the negative of (-h)(5). Using the negative of one factor produces the negative of the entire quantity. What is the negative of (~3)(-#)(-2)7 §-2;§-;2 = -5 7 Fill in the blanks with the negatives of the given quantities. Check the answers given in the last problem if you haven't done so before. Note that both of them give the same value and this is the same value as the original So far we have: Now consider §2%§%% . Is the value of this equal to the value of the fractions above? How do you know? In this case, we didn't use the negative of both the numerator and denominator. ‘We changed the signs of two factors or we could say that we multiplied two factors by -l. Multiplying two factors by -l keeps the same value as multiplying two factors by -1 is equivalent to multiplying by +1. Multiplying by +1 doesn't change the value. For example, (-2)(3) = (-2°-l)(3’-l) = (2)(-3). write four different expressions which are equivalent to é-4;§§§ by changing signs. -3 7 Remember we always multiply two factors by -1 to keep the same value. Can we write that fi:§ _‘:512 Why? 5-2 ‘ -5+2 ? M -151... 29. They are equal if (4~2)(-5+2) 30. = (“'1‘th) (5‘2) 0 This becomes (2)(-3) = (~2)(3) which is a true statement. The given statement is correct. 30. No, it is a term. Factors are 31. associated with multiplication. 31. negative of 5-2 is -1(5-2) or 32. -542, h-2_.LH-2=:g__._g=,2 5-2 - -5+2 3 -3 3 320 2 = “'2 "' '1'" 330 {-5+3§E-35 5-3 3 = 2 - h _ -2 ~5+3 3 5-3 -3 = 2 - -h = -2 és-i 352-3; é-fiBEEB; Any two of these are the answers to frame 32. 33. Yes because (-7b2)(-#a2) = (7b2)(ha2) or because the numerators and denmminators of the two fractions are the negatives of each other or because two factors in the fraction have been multiplied by -1. numerator is -2pq(2p-5q) or is 2Pq(-2p+5q). As long as two factors are multiplied by -1, then the value remains the same. In this case, we multiplied one factor of the denominator by -1 so we must multiply one factor of the numerator by -1 to have equivalent fractions. 35. C 2 .- a+b aWb = bZ2a+bg fpi2§+3q§ = You may be saying that too many signs were changed. However, remember that two factors must be multiplied by -1 to keep the same value. In. fi=§ is 4 a factor? 5-2’ Then to find the negative of (4-2) we must find the value of -l(4.2) which is -4+2. The negative of 5-2 is . So, 2:; _ —#+2 5-2 - ..2 ....u = = -5+3 -3 --—- Does :fli equal .293 ? Why? haz -4a2 In other words, if we multiply two factors by -l, we obtain an equivalent fraction. ? p(2p+5q) 35. write two other fractions which are equal to the given fraction and make use of changing the appropriate signs. a2 a-b a+hb b 2a+b 36. Note the answers given to the last problem. Any of these are - 152 - - a2 a-b ~a-hb 2 correct. There are also thggg b 2a+b more possible answers. What are 3aZL-3a+b)L-a-I+m; 3.29.1-1.) (swab) these? b(2a+b) b(-2a-b5' _- az a-b a+#b _ -3a2§3a-b)(a+4b) - -b(2a+b5 b(-2an) _ 3.21.3.4.» Lawb) ’ -b(2a+b) 36. 3e2(-3e+b)(e+4h) = 37. We do not multiply factors by -l b(-2a-b) in every case. This is only done a2 a-b -a.4b = 3e2(3a-b)(-e-ub) when it will help us in handling -b 2a+b EZLZa-b) the quantities. In the fraction a- , note that éB-ag all the signs of the factor in the numerator are the opposite of those of the factor in the denom. inator. In other words, the numerator is the negative of the denominator or the denominator is the negative of the numerator. If we multiply the factor of the numerator by -l, we get -a+3. This is exactly the same as the factor in the denominator. However, we have only multiplied one factor by -1, so we do not have the same value. we must multiply another factor by -1 in order to have a value equal to the original. éa-fig = S-a+22 . Now the 3-a -1 3—a two fractions have the same value. Let a = 2 and evaluate both fractions. What values do you get? Are these equal? 37. a- = 2-“ = :l = l 38. Consider -a+ . Both the é3—a; 3-2 1 ‘ -1 3-a -a+ = ~2+ = l = -l numerator and denominator have -1 3-a -1 3-2 -lZl5 a common factor which is . Dividing both numerator and denominator by this common factor, we get . 38. common factor is -a+3 or 3-a. 39. Consider the fraction Note that -a+3 and 3-a are the .22%2191i§%:2317 . same factor. -9 3p-2q -p-q Dividing both numerator and If we are to reduce this fraction -153- denominator by ~a+3, we get l/-l which equals -1. 39. common factor is 3 no, 40. $339252?ng ul. -3 3p-2q -p-q 41. yes, p+q and -p-q are negatives 42. of each other. That is, p+q = —l(-p-q) and -p-q = -1(p+q). to its lowest terms, we must divide both the numerator and denominator by the same number or we must divide both numerator and denom. inator by a common factor. As the fraction is given, what common factor is found in both numerator and denominator? Divide both numerator and denom. .inator by this common factor and write the result. Look at this result. W -3 3p-2q -p-q Are there any factors which are the negatives of each other? If so, what are they? If two factors are the negatives of each other, it is to our advantage to change the appropriate signs so that the same factor appears in both the numerator and denominator. Write a fraction equivalent to Bé2192§2?:292 -3 3p-2q -p-q having the factor -p—q in both numerator and denominator. 42. pé-p:92§22:39; or'2§‘2292§22=29; #3. In each of the possible answers 3 3p-2q -peq -3 3p-2q -p-q to the last frame, only two - -2 - 2 ° ~3z3p-2q5z-p-q; ongénggg駧%§%%§7 “3. Bégp:2q§ or :3égp-2q; or “a. factors were multiplied by -l. Go back and check to make sure that this is true. Now you have a fraction where the numerator and denominator have a common factor. ‘Write an equivalent fraction in its lowest terms. Any of these are correct, so therefore they must all be equivalent. By assigning values to p and q, you can check to make sure that the fractions are equivalent. Remember, don't assign values to p and q which will make the denominator equal to 0. Why can't the denominator equal 0? 44. division by 0 is undefined 45. 46. “70 2X XOZ BZx-ZS - 15h - 45. or can't divide by O (a) = 23a£a+2) -a3e(e+2) numerator and denominator have common factors of -a(a+2) and so equals fi 2 O b (b) = a-b a+b -b a-b numerator and denominator have common factors of a-b and so equals 3a b b O (c) = §x~12§2x-212 Now divide -1 xey 3y-2x both numerator and denominator by the common factor of x-y, getting ZX-Ex . -1 3y52x Now notice that 2x~3y and 3y-2x are also negatives of each other. Then, the above fraction equals ZxfiZZ and dividing both +1 2x-3y numerator and denominator by the common factor, we get l/l or 1. Now both Now both factor the numerator and denominator 46. 47. 48. Express the following as equivalent fractions in their lowest terms. (a) -§a$-a-22 3313(84'2) (b) (a-b)(3a+bl In these last problems, you may have changed different signs from those indicated in the answers. It makes no difference which signs you changed as long as you multiplied two factors by -1. Check and make sure that you have multiplied two factors by -1. You will note in part c, four factors were multiplied by -l. The value of the fraction will remain the same if an even number of factors are multiplied by -l as the total effect is that of multiplying by +1. We have been careful to mention that both numerator and denomp inator must be divided by a common factor. If we are to reduce the following fraction to lowest terms, we don't have factors, we have terms. 2x2 - 4x What could we do in 3x - 3 order to find out whether the numerator and denominator have any common factors? Factoring the numerator and denominator, 2x2 - 4x = 3x - 3 Now the numerator and denominator of this fraction have a common factor of x-2 and dividing both numerator and denominator by this factor, we get . - 155 - 48. 2;, This fraction is in its 3 lowest terms. 49. Factor both numerator and denominator getting 39.2% a-b) g2e+3b2 a 2a+3b DiVide both numerator and 49. 50. denominator by the common factor of a(2a+3b) getting 3a(a-b). 50. (a) éxefiééx—Z; =‘§:g x—3 x+l x+l (b)ééfl3%§:lg.=§12 c+d xey c+d (C) flflx-sz = 2x2S inc-g): 3xy2<2y-3x) 2Y-3x W = -3x2 2y-3x Don't forget about multiplying two factors by -l in order to keep the same value. (d) gxezzgxz+2x+uz = x2+2x+4 xez xiZ x+2 (e) £4x3-108zgx2-12 z 8x(l~x)(x2+3x+9) 51. Reduce to lowest terms and state the procedure you used. 6au+3a3b-9a2b2 2a2+3ab Factoring a polynomial always means to express as prime factors. Express each of the following as equivalent fractions in their lowest terms. (a) x2-§x+6 x2-2x.3 (b) axeay+bx~by cxecyvdx-dy (c) 22x4y2-18x3x3 6xy3-9x2y2 29.2.8.3. x2-4 4x5-4x3-108x2+108 (8x-8x2)(x3+3x+9) 24a5b6-36a6b5 l8a8b6 (g) 120x7x9z2 75x5y622 (h) 2ab- cb-4ad+20d ( i) (d) (e) (f) 2cb.4ab+cd-2ad x+2 xfil xil x-Z Let us examine the fraction in part i of the last frame again. You were given éxngxrl . xil x-2 The numerator of (x-2)x+1 tells you to multiply (er) by x and then add 1. If you were told to multiply (x52) by'(x*l), it would be written (x-2)(x+l). Notice that this is pat the same as what you were given. In order to reduce a fraction to its lowest terms, both numerator and denominator must be divided ~156- by the common factor. ugx.32§x2+3x+22gx.lzgx+l) = 8x(l-x)(x2+3x+9) Is xp2 a factor of (x-2)x+l? 4 x- x-l x+l = 4(x—2)(x—l)§x+l) 8x l-x -8x x-l In the denominator (x+l)x>2, is = (x-fizgxml) xml a factor? -2x You might not have exactly the Is x-2 a factor of the denominator? same answer, just make sure it is equivalent to this one. If you are in doubt, substitute a numerical value for x. (r) 6a5b5(4b—6a) = 4b-6a l8a8b6 3a3b (g) divide both numerator and denominator by 15x5y322 as these are already factors. Answer is 8x2/5y3 (h) b(2a-c)-2d§2a-c; = 2b c-2a +d c-2a Is x+l a factor of the numerator? éb-Zd;é2a-c; = -b+2d ~2a+c 2b+d c-2a 2b+d c-Za -b+2d =2b+d (i) WARNING (x-2) is £22 a factor of the whole numerator. (x-2) is only a factor of the first term of the numerator. x2-2x+l __, éx-lgéx-lg = 3:; x2:;L2 x+2 x-l x+2 51. The answer is no to all of these 52. Express each of the following questions. For x+l to be a factor of the denominator, it must be multiplied by every- thing else in the denominator. This is true of any factor. In (xil)x-2, x+1 is a factor of ongz the first term, pat of the complete denominator. 52. (a) 2x2.,+3x+1 = 2x+1 2x+3 x+l 2x+3 xfil as equivalent fractions in their lowest terms. ( ) §2x+3§x+l a 2x+3 x-I-l (b) (x~§)(x+2) x2+2x (c) éfip=§;(p-2) 3p-5 p-2 (d) a6-b6 a“-b” (e) (fix2+2§y.§zEZS§:X2 (3x?+8xy+5y2)(xey) 53. If we were to multiply 2/9 by 12/16 or g o 12 using the law 9'1'3’ _ 157 - 2_ 2 = 3;”1 3p 5p- _ 32 E (d) L the +a2bg§gfl z (a 2-b2)(a2+b2) au+a2b2+bu a2+b2 (9) xi x- x+ 3x+5y x+y X-y Note - the answer is not 0. You are dividing a number by itself and the result is 1. 53°§Z'§§=l°2=% 8 81 2 3 54. (a) Remember 32 is an integer, and any integer can be written with a denominator of 1. 1°21’3_2.-. 3 33' 1 14 (b) Product is 1. both numerator and denominator are divided by a factor which is the same as the numerator and denominator or when a number is divided by itself, the result is 1. 55. factor the polynomials Remember when for the multiplication of frac- tions, we would get 2°12 = 24 9-18 TIE? This fraction is not in its lowest terms. To put it in its lowest terms, we must divide both numerator and denominator by their common factors. This means dividing both numerator and denominator by 24 getting 1/6. Since we can always get equiv- alent fractions by dividing both numerator and denominator by a common factor, we could do this before we multiply. 3 2 This is the same Eresult as when3 24/144 is reduced to lowest terms. For example, 2'12_:;_g-1'1= Multiply: 27/8 by 28/81. Remove common factors before you multiply. Find the following products. ress in lowest terms. (a ‘1 ° 21 ' 3 i3’ 32 (b) 8 ' °‘_Z ate When the numerator and denom. inator of a fraction are polynomials, we can divide by the common factors and then multiply or multiply and then reduce to lowest terms. Usually the first procedure gives us lower degree polynomials to deal Withe For example, x2-2x ° 2x2-Zx+§ x2-5x+6 x3-4x2 Before we can reduce to lowest terms, we must first . Factor these polynomials, divide both numerator and denominator by their common factors and multiply the resulting two fractions. x2-2x - 2x2-zx+3 = xz-5x+6 x .4 -158- 57. Find the following products. 56. x x.2 ' {ZXhlzgx-fiz x—2 x-3 x2(x.n) = 2xpl x x- 57. 580 Both numerator and denominator can be divided by x(x—2)(x>3). (a) :ZfabBCZ ° (2a+3) (a—8) 58. (a-8)(a+8) Slaub7c = 2c ° 2a+3 = 20§2a+3) (a+8) 3a3bE 3a3bu(a+8) (b) (21-4) (azwaflél ' aux-4) (a-fl)(a~#) a(a2+4a+16) = l (c) 6- x 6+ x ‘ x-Z 5X- x+l (5;;ZSZX-ZSZX+25 = 6-5x and now (5x~6)(x#1)(x+2) multiplying two factors by -1, this becomes é-6+gxg _ l 5xp -xpl x92 - (-x-l§(x#27 There are other possible answers here. Just make sure that any other is equivalent to this one. (a) 2 5xy222 (b) l (c) }215 Don't forget to multiply -p two factors by -l. (d) 2 -l x+l éxflSEZXfS; (e) Sx-iiZZQX-l-lz ( -7x~6)(x-l) don't forget, x-l 15,223 a factor of the first numerator. (f) 1 (g) - S2x§lZS2x*§2 x-l or equivalent. 59. Express results in lowest terms. (a) 34ab3c2 ' ZaZ—IEa-24 a ~6H 51a 70 (b) 2.3.194...- ° aZ-ha a2-8a+16 a3+4a2+l6a (c) fi6—22x2 ' x-Z :7_ 5x2-xa6 5x3+6x2-20x—2# (a) QEXBXEZ ° 48x7 823 = 55x6y9zg 72x5y3 (b) a+6b - M = a+8b 6a-9b _2?;_16_-223_+22=2= 4-3p-p2 pz-up (d) 2x2+xsl - x?-u . 3x2+2x.21 3x2-13x+1n x2+x-2 2x2+11x+15 (e) xgx-lz-QSxplz+2 ’ sz-fix-fi = (x—l)(x—2) x2-7x-6 (r) x3+64 ' x3-x&10 ° x3y+4x2y y3+2y2+4y+8 x213+4fi .- x2(2y-5)+(16-4x) (2y-5) (g) M'uxzflzmy £41.15 (ZX-5)(xel) 9-x2 2x+1 Notice that the answers to the problems in the last frame are in factored form. You can multiply the factors together or you can leave the answers in factored form. Addition of fractions. If b O and a, b and c are real numbers, than 9'. £=l. +11. =35 zét—c- b + b b a b c be<:x21izimfiz-_ac+_zi.° XV x2-xy+y2 ( x31?( 7 3 3x+ x+ x212 _ x2 2 2y y l - y <61_2$2:21_ - [lap-551:9 gong-n+9} =.212:22_.-$£:£D__.-.1L. f-3p+9 p2-3p+9 P'3 a-b 2a- b+2a b—2a m§3a-b;(2:-b) 3b-2a 3b+2a _ 1 (Don't forget to multiply two factors by .1.)a (g) @a 2-8a+§ 3a+l a-3 a-7a-3 a—l 2a-6 a+l ( 3a+l;(a-3; 3a—5 a-l 2a- a-3 (h) This is addition so you must get an LCD. LCD is c(c+d). (c-d)(c+d)- cd.d2 = c.2d c(c+d) (1) LCD is x(x-3)(x2+3x+9) x2(x-3)+(x-2)(x2‘t3rt9)-x = x(x-3)(x2+3x+9) 2x3-2x2+2x-18 x(x—3)(x2+3x+9) (j) LCD is (a-4)(a+3)(a-3). a a- + a.u - a+l a+ _ a a+3 a-3 -#a-l5 (aéh)(a+3)(a-3) 102. 2a _733 - z 1 ;§_ ' 1 'l5 '73 5 2. —2—-=-2—.l-8.=-L£ €g, 3 7 9 103. 4 by 22 103. 104. Given the fraction, 2+4 we 22 could write it as (2+4); (22.4). We can't write it as 2+4- ' 22-#. This last statement tells us to diVide by e 2&4 22 to divide.________'by Given the fraction, it means and 104. 105. 106. 107. 108. 109. - 17o - (2+4) by (22.4) 105. simplified to 6 divided by 18 or to 1 divided by 3 (2W) -;- (égfl) or Q = g 106. 9 3 x2+2 107. X x 1+5 x2+2b£+£ 108. l 2 Notice that this is pat the same as the answer to the last frame. First combine the terms in the 109. numerator and denominator. This equals (2 + g) .:. (% + 1) or equals 11:21-22. "3'8 9 (g - a) -: (9-163(2) = M.‘ l =1 x (3:5x)(§:5x) x(3;5x) 110. this can be simplified to read diflde by 0 Write (2+4) 22.4 2 as one number divided by another and simplify. Write (x2+ 2) :. (1+ X) x ' '2 in fractional form. DO NOT simplify. Write another expression equivalent to x2+211+ Ft» Aux Simplify: 1 5+1 In this problem, you must first do what Operation? If you can't decide, rewrite the expression as something divided by something. Do the simplifying. l-u X the proper places. 9-16::2 In the case of any complex frac. tion, the numerator and denominator must be expressed as one fraction before any further simplification can be done. You may express it as one quantity divided by another quantity if you want to. Be sure to use parentheses in If a problem is given with parentheses in it, be sure and pay attention to the signs of grouping and what they signify. Perform the indicated operations am simplify o ‘a’ <2-§*;%>€-<1--%> X 110. (a) ($29.2) i (23:3) 2 x2 ‘ x2 2 -1 -2 Eiffel (x-2)(x+27 (b).§12£i53.: x2 0 g2¢§2g2+x2 ° x = x2 3x+2 +x 2+x x 3x+2 (0) 2:37.]; .3. 30-241 = x3 ' x-2 (xel)(x2+x#l) ' er - x-l 111. gx=22gxe§2-18 ;_x2-16 = X-S x?-2x.1o-1 ' x-5 figx.2 ' X-5 (xezzgx+uz ’ X-S X- X‘I’ X 7N a W. X+ = %%%= W: ’°’ = __e:.,r: — v.77 7 __.——_ -171- E? v Nlox ... .... H x» x ... Nina Nlm A O v ....I I H “uh-a 1+ x-2 111. In part c of the last frame, the denominator became _)_c_-_§_'_+_l_ x—Z Note that we must combine similar terms before we can decide what factor we have and so we write _2x-1 112. Note the numerator of the last fraction. In going from the second to the third step, we had to combine similar terms to get x2-3xe28 before factoring. You will often find that you don't do the steps exactly in the same order as in the answer, but be sure yours are equivalent. You will often find some steps are missing in the answers, but you ought to be able to supply these by this time. 2b a-b l a+b 2 a-b - 172 - 112. 2b-(a-b2‘_ 2(a-b)-(a+b) = a-b a-b fib—a ‘ a-b = -l a—b a-3b Remember to multiply two factors by -1. 113. (a) (a2+b2) (a-b) (m) 1m) ($132) 2 a(b+¢) (e2+b2) (3-13) 3 (b) 6x3y“?x(x-2y)(x+2y) 27x5y2 z aflmzy) 9::(x2-o-2xy'l'43'2 (c) -§c+a2 Two factors must 3 be multiplied by -l. [4 (6) (mz-(fix-y) = -2x+21 = _2 xsy x-y (f) -b +b +b _ 2+1)? W ‘ (a-b)(a+b) (g) l-xfixSl—x2-1+x2 g 0 l-x 113. Perform the indicated operations and simplify. (a) an-bn ° aZb-b3-l-a2c-cb2 a2 ab+ac (b) 6x31& ° x3-4§y? x.3--8y'3 27xzsy2 (c) 312 ° cz-a2 a-c 3a+3c (d) (x-yzu‘_ xz-Zzyzyf BX-Y 6x2+xyey2 (e)£:z_2.x_-x xey xpy (f)..__e__ + .10.. a+b aeb (g) .125. ...}:53 Lot.... X 1.x (h) ..2_ __._1 +_———4 (i) 2x2+3x.12 + _l_ x3—4x2-9x+36 3-x a+b (k) 8a-l 3a +«—;:§ _ 2a-1 33 a+2 -173- 2m—l 2m+1 ' (2m-l)(2m§l) (i) 2x2+3xe12-gxe42gx+22 = x x-3 x+3 x2+ux (XIWx-3)(x+3$ (3) 21a ° ..L .. .1. b a+b - b (k) 3a(a-2)+8a-l ° a+2 = a-2 3a(a+2)-(2a-l) a-l a+2 (a-Z)(3a+l; Chapter 6 - Linear and Fractional Equations This chapter deals with the.solution of first degree equations. We will also be concerned with analyzing word statements, translating these word statements into symbols, obtaining equations which express relationships given in the word statements and solving these equations. Knowledge of the techniques and skills presented in the first five chapters is necessary before proceeding with this chapter. 1. 1. because if x = 3 the denominator 2. would equal 0 and division by 0 isn't defined An equation is a statement of equality between two quantities. x+2y = x+3yey is an example of an equation. This equation is called an identity because it is true for all values of x and y. We have some identities which are true for all values of the variable with one or two exceptions. For example, ac. .2. .. 211. x-3 x-3 - x-3 is true for all values of x except 3. Why can’t x = 3? We are going to be dealing with the solution of conditional equations. These are equations which are true for only one, two or some finite number of values. They impose a condition or conditions on the variable. x - 2 = 5 is a conditional equation. There is only one value of x which makes this statement true. Actually we can substitute any real number in place of the variable, but if we let x take on any value other than 7, we get a false statement. For example, if x = 3 then x-2=5 becomes 3-2=5 which is 222 true. x-2=5 is an example of a linear equation. A linear equation is an equation in which the variable appears only to the first power after similar terms are combined. Which of the following are linear equations? (a) 2(X-2) + 7 = (b) x(x+2) - 3 = 0 .. 174- - I75 - ‘°’-’£‘§-E=% (d) (2x-l)(2x+3) .. (42624630 = o 2. a, c and d are linear equations. 3. Two equations are equivalent if 5. Be sure you did the multiplication and combined similar terms before you decided. While it isn't necessary in part c to get the LCD before you decided whether or not the equation was a linear one, you must get the LCD in more complex problems. linear 4. root equivalent 5 . Remember - equivalent equations must have exactly the same solution or solutions. one they have the same solutions. It may be verified by direct substitution that a=2 is a solution of 2a=Q and of 3a-6=0. Thus 2 is a solution for both equations. A solution to an equation is called the root of the equation. 3a-6=0 is a equation and 2 is a of this equation. In the last frame, 2 was a root of both the equation 2a=4 and of 3a-6=0. If 2 is the only solution of these two equations or if some other number is also the solution of both equations, then these equations are equivalent equations. 2a=4 and 3a-6=O are both linear equations and linear equations have at most one solution or root. Thus, 2a=h and 3a-6=0 are e equations. In each of the following, you are given two equations and a value of the variable. These are all linear equations so have at most root. Determine by substitution if the given value of the variable is a root of the equation and state whether the two equations are equivalent equations. (a) 3x-2 = 2x+l; “(x-3) = 0; x=3. (b) x(x—l) - (x-2)(x+3) = 2; x _ . x = 2. xe2 + 2 - x, (c) 4x = x+93 9x.# = 3x-22; x=3. - 176 - 6. only the equations in part a are 7. In solving equations, it is often 7. 9. equivalent equations (b) 2 is a root of the first equation, but not of the second one. In the second one 2 makes the first denominator equal to O and division by O is undefined. (c) 3 is a root of the first equation but not of the second one. yes because the same number, -2, 8. has been added to both sides of the equation or because 2 has been subtracted from both.sides of the equation. convenient to find an equivalent equation, so we must be able to determine when we have an equiv- alent equation. Equivalent equations. 1. An equation equivalent to the given equation is obtained when the same quantity is added to or subtracted from both sides of the equation. 2. An equation equivalent to the given equation is obtained when both sides of the equation are multiplied or divided by the same nonzero number. Given the equation x+2 = 5, is x = 3 equivalent to it? Why? Given the equation 6xp18 = 15, is 2x96 = 5 equivalent to it? Why? yes because both sides of the 9. Given the equation‘_l equation were divided by 3. 2x = 3 is l = 6x equivalent to it? Why? It is £93 equivalent if x = O, ale. Given the equation l-x = 4, is then the denominator of the left side of the original equation is 0 and a fraction with 0 in the denominator isn't defined. These two equations gag equiv- alent for all values of x except x = 0 because both sides of the equation have been multiplied by the same nonzero number. the equation 4.4x = 16 equivalent to it? Why? (b) is (1-x)(2+x) = 4(2+x) equivalent to it? Why? (0) is %.- x = 2 equivalent to it? Why? ((1) is lg = _L equivalent x+2 x+2 to it? Why? 10. (a) yes because both sides have 11. In solving linear equations, we been multiplied.by the nonzero number of 4. (b) yes if x f -2 because then both sides of the equation have been multiplied by the nonzero quantity of x+2. change the given equation to an equivalent equation until we have arranged the equation so that all the terms containing the variable are on one side and all other terms are on the other side. -177- (c)‘qq‘because the complete left Then we continue changing to side of the equation hasn't been equivalent equations until we divided by 2. get 1 multiplied by the variable (d) yes if x f -2 because than equal to a number. This number both sides of the equation have is the root of the equation pro— been divided by the nonzero vided substitution of this value quantity of x+2. makes the original equation true. For example, to solve 5(x-2) + 1 = 3x+?, we could first remove the parentheses, getting 5x-10+1 = 3x+7. Adding -3x to both sides of this equation, we get the equivalent equation 2x—lO+l = 7. Adding 9 or subtracting -9 from both sides of this equation, we get the equivalent equation 2x=16. Dividing both sides of this equation by 2, we get the equiv- alent equation x = 8. Substituting 8 for x in the original equation gives a true statement, so x = 8 is a root of the equation 5(x92) + l = 3x+7. Notice that in the solution given above, we have equivalent equations in all cases since we added the same quantity to both sides of the equation and divided both sides of the equation by the same nonzero constant. Using the equation 5(x-2) + l = 3x+7 find equivalent equations by first adding -1 to both sides of the equation and then by dividing both sides of the equation by 5. 11. 5(x-2) + l = 3x+7 12. Take the last equation obtained Adding «l to both sides, we get to frame ll and find equivalent 5(X92) = 3x+6. equations by first adding':3y Dividing both sides by 5, we get 5 x—Z =‘2§ +‘é to both sides and then by adding 5 5 2 to both sides of the equation. 12. Adding :% x to both sides, we 13. we now have the equation 5; x _ _Jé 5 - 5 get .2. x _ 2 = 5:3,. or 235 .___, _lé What two processes 5 5 5 5. Adding 2 to both sides, we get can we do to find out the value .2 x =.l§. of x? Will you obtain equivalent 5 5 equations in each case? Why? - 178 - 13. Divide both sides of the equation 14. Perform these two operations 14. 15. by 2 and multiply both sides of the equation by 5. (These opera- tions can be done in either order.) Yes, equivalent equations are obtained in both cases because multiplying and dividing both sides of an equation by the same nonzero quantity gives equivalent equations. Multiplying both sides by 5 giveslS. 2x = 16. Dividing both sides by 2 gives x = 8. x = 8 is a root because when it is substituted in the original equation, a true statement is Obtained o It makes no difference the order 16. in which operations are done as long as the same operation is done to both sides of the equation and equivalent equations are obtained. and obtain the equivalent equations. Is the value obtained for x a root to the equation? Why? This root of the equation and any root of the equation can always be checked by direct substitution in the original equation. When the value of the variable is substituted in the original equation, a true statement must be obtained. The equation 5(xp2)+l=3x#7 was solved in frame 11 by obtaining certain equivalent equations. In frames 11 - 14, you were told to find other equivalent equations starting with this same equation. Note that the same root was obtained in both cases. What can you say about solving equations by obtaining equivalent equations. Find the root of the following equations and check your result. In parts a and b state the pro- cess you used to obtain equivalent equations. (a) 3a~4 = 5a+6 (b)2E—3=-]2-1x+1 (c) 3(xs4) - 4(2x+l) = 4 (d) (ZX-l)(3x+2)-(6X*l)(Xb3) = 13 E2; gp(§p+3)8- iip-Z) = 6p2-2(p+3) x- Z x+ _ x 3 " 3 ”‘23 (Ql_2:5 Y Y ezi_e=a=_i_o a-3 a+3 a2_9 (j)_35_3§__.g_.b_ solve for x. b a b a ’ (k) a: .. _._...2a2-2 - .. 2:2 3+1 2 83-8 (1) 2m....1 _aal'iafl m+2 m~2 (m) x-2 l8 - 182 - 24. DON'T FORGET to do the checks for each problem. (a) x = 15 (b) a = l (c) p = -60 (d) p = 20 (e)y = PEZX (f) r =‘qA 21f (g) b = -Sa+22 a (h) _ -d-c x - c—d (i) no solution (3') x = a+b (k) a = 5 (l) m = 3 (m) x = 2 25. (a) t = rs Sir - r (b) S _ t-r Parts a and b are true only if r, s and t ¥ 0. Also, in part a, s and r can't have the same value. t and r can't have the same In part b, 250 26. 'value. Why? (0) W'= -14/3 (d) x = ll/2 (e) no solution (f) R = -Wr only if P and W Paw aren't equal (g) x = -(c+d) if c and d aren't 0 and aren't equal. (h) a = 6b if b # -3/2. -2b-3 Solve and check the following equations. (a) l_l+_l_ solve for t. " 9 r s t (b) Solve the equation in part a for s. (c) w' _’w+ _ *“ mini w2-16 (d) .2_.._.§..=—i—— x2-5x+6 x2-4 x2-xe6 (e) x+2 xel 2x2-3x 2x2-5x+3 (r)P=¥-§%=’-‘l,rora. ..l - x (9.25-2: c d c d d c (h) ab-3(a+2b) = 3ab, solve for a. for x. Make sure you understand why the limitations on the values of certain variables apply in the last frame. You.may never have a denominator of 0 as division by 0 is not defined. This condition applies to the given equation and to the solution. If you are givenwi = 3, even though it is not stated that x aé o, it is understood that this condition applies. We can never use any value of a variable which gives a denominator of 0. The main purpose of learning to handle algebraic expressions is to solve equations. In.many cases, we will have a relationship stated in words, and we must translate this into symbols and obtain an equation expressing this relationship and then solve the equation. The main problem in translating word statements into an algebraic statements seems to be in recognizing what the symbols stand for. Therefore, we will Spend some time in translating word 26. o o‘m 9.. VV VVV AA AAA . 183 - 27. SIS 5 ' ' a" 3 n - m (n-5)2 or 2(n-5) NOT n—5‘2. This says n minus the product of 5 and 2. NOT 2’n—5, this says only 2 multiplied by n and then to subtract 5. 2n+6 or 6+2n 2(m-n) or (m-n)( 2) 3(n-5) or (n-5)(3) statements into algebraic symbols. If we were to write the sum of two numbers, we would first have to know what numbers we had. Thus to represent the sum of two numbers, you first must represent the two numbers. If you are just told that you have two numbers, they may be represented as follows. Let n = one number Let m = the other number You will notice here that we chose two different symbols to represent the two numbers as we don't know what relationship holds between the two numbers. Now we can use n+m to represent the sum of the two numbers. Let m and n represent two numbers. Represent (a) the product of m and n. (b) the difference of m and n. (c) the quotient of m and n. (d) subtract m from n. (e) the product of a number 5 less than n and 2. (f) a number 6 more than twice n. (g) a number which is twice the difference of m and n. (h) a number which is three times the difference of n and 5. If you are not sure of the relationship expressed when letters are used, state the problem using numbers. Decide what process you used when the numbers were in the problem and then do the same_process using the appropriate letters. For example, If a car goes m miles per hour, how far does it go in h hours? This is the same type problem as "A car goes 40 miles per hour, how far does it go in 3 hours? 26. (a) (b) (C) (d) (e) _ 183 - mn 27. m - n .9 n n — m (n-5)2 or 2(n-5) NOT n—5‘2. This says n minus the product of 5 and 2. NOT 2°n-5, this says only 2 multiplied by n and then to subtract 5. 2n+6 or 6+2n 2(m—n) or (m-n)(2) 3(n-5) or (n-5)(3) statements into algebraic symbols. If we were to write the sum of two numbers, we would first have to know what numbers we had. Thus to represent the sum of two numbers, you first must represent the two numbers. If you are just told that you have two numbers, they may be represented as follows. Let n one number Let m the other number You will notice here that we chose two different symbols to represent the two numbers as we don't know what relationship holds between the two numbers. Now we can use n+m to represent the sum of the two numbers. Let m and n represent two numbers. Represent (a) the product of m and n. (b) the difference of m and n. (c) the quotient of m and n. (d) subtract m from n. (e) the product of a number 5 less than n and 2. (f) a number 6 more than twice n. (g) a number which is twice the difference of m and n. (h) a number which is three times the difference of n and 5. If you are not sure of the relationship expressed when letters are used, state the problem using numbers. Decide what_process you used when the numbers were in the problem and then do the same process using the appropriate letters. For example, If a car goes m miles per hour, how far does it go in h hours? This is the same type problem as "A car goes 40 miles per hour, how far does it go in 3 hours? 27. 28. 29. 30. 31. 32. 33. - 184 - distance = rate per hour multi- 28. plied by time in hours. multiply m which is the hourly 29. rate by h which is the number of hours. The distance is mh miles. Multiply x which is the hourly 30. rate by 1 1/2 hours or‘3 hours 2 which is the number of hours traveled. The distance is 2% miles orig x miles. Change the time from minutes to 31. hours because the rate is given as miles per hour. .2 l 32- 20r2y You multiply the number of 33. pounds by the cost per pound. Since the cost is given in cents, ab represents the cost in cents of the coffee. To change 35 cents or 190 cents 34. to dollars, you would divide by 100 since this is the number of Decide what process is necessary in order to solve the second problem - don't worry about what the answer is. What relationship would you use to solve the second problem? So we would multiply 40 miles per hour by 3 hours or we would multiply the rate by the time. Use the same principle to represent the number of miles traveled if a car travels m miles per hour for h hours. If you travel at x miles per hour for 1 hour and 30 minutes, represent the distance traveled. Suppose you wanted the distance traveled when you travel at y miles per hour for 30 minutes. What would you have to do before you multiplied the rate by the time? Represent the distance traveled when you travel y miles per hour for 30 minutes. Represent the cost of a pounds of coffee which sells bor b cents a pound. Represent this cost in cents. Suppose you needed the cost expressed in dollars of a pounds which sells for b cents a pound. The result to frame 32 which is ab cents would have to be expressed in dollars. Decide‘hqy'you would express 35 cents or 190 cents in dollars and then do the same process and express ab cents in dollars. You could express‘_qb in decimal 100 form instead of in fractional form. -l85- cents in one dollar. NOTE - determine the process not the result. Now Change ab cents to dollars using the same process. 18% dollars 3“. (a) lOOd cents (b) .22 100 (c) a(x—3) or (xe3)a cents (d) _gy or .ley dollars. 100 (e) you need the cost of one pound of sugar which is 25 3 cents. So the cost in cents of w pounds is 223 or ggyw 3 3 ' cost in dollars of one rose or .O3x dollars (f) is d/12. Thus, the cost of x roses is _d_ _dgg 12 x °r 12 dollars. (g) The time needs to be ex. pressed in hours. m minutes equals m/6O hours. 35. ab 1 and l _._.=.___ b ———-= , 100 100 a 100 01 8° ab _._.= , b 100 Ola You cannot write .ab as this has no meaning. A decimal point only has meaning when the number is expressed with numerical digits. Represent each of the following. (a) the number of cents in d dollars. (b) the number of dollars in (3x) cents. (c) the cost in cents of a pencils at (xp3) cents each. (d) the cost in dollars of x pounds of sugar at y cents per pound. the cost in cents of w pounds of sugar which sells at 3 pounds for 25 cents. the cost of x roses which sell at d dollars a dozen. the number of miles traveled by a car which travels p miles per hour for m minutes. the rate of a car which travels 80 miles in x hours. the number of hours traveled when a car goes d miles at x miles per hour. (9) (f) (g) (h) (i) 3,4,5,6 are examples of consecutive numbers. Notice that one is added to each number in order to get the next consecutive number. 3, 5, 7, 9 are examples of consecutive odd numbers. 2, , , 8 are examples of consecutive even numbers. Consecutive odd and consecutive even numbers differ by 2 or in other words, add 2 to a number in order to get the next consecutive odd or consecutive even number. The resulting number is odd or even depending on whether the original number is odd or even. 35. 36. 37. 38. - 186 - distance is 3% p or ppm-165. (h) We know that distance rate multiplied by time. Here we know the distance and the time, so how would we find the rate? rate In this problem, r =-%% . (i) number of hours = Nip. sometimes 36. n+2 is even if n is even. n+2 is odd if n is odd. If n+3 is even, than n must be 37. odd. n+4 is the next higher consecutive number. n+3 represents an even number so n+4 represents an odd number. n+2 and n+1 are the two consecutive numbers which precede n+3. Remember consecutive numbers differ by l. 38. n+1 and n-1 are the consecutive even numbers which immediately precede n+3. Remember consecutive even numbers differ by 2. (a) w - 7 39. (b) W'+ 11. w + 4 represents Susan's age now. (e) 304 + 7) Does n+2 represent an even number sometimes or does it always represent an even number? Give a reason for your answer. distance divided by time. If n+3 represents an even number, what can you say about the value of n? Write the next higher consecutive number. Is this an odd or even number? If n+3 represents an even number, represent the two consecutive numbers which immediately precede n+3. Represent the two consecutive even numbers which immediately precede n+3. If w represents John's age now Susan is 4 years older than John, represent (a) John's age 7 years ago. (b) Susan's age in 7 years. (c) three times Susan's age in 3 years. Represent the following. (a) The sum of two numbers is 7. One of the numbers is y, what is the other one? The product of two numbers is 12. One of the numbers is w, what is the other one? (c) One number is t. Another number is 5 more than twice that number. The second number is . (b) 390 (a) 7 " y (b) 12/w (c) 2t+5 or 5+2t 40. 2n+5 3n-3. NOT 3-3n. 41. These quantities are equal. 2n+5 = 311-30 42. the required number is 8. -187- 40. When finding a value which 41. 42. 43. satisfies a statement problem, we must first discover a relationship which holds between the quantities we have represented. For example, given the following. Five more than twice a number is three less than three times a number. Find the number. First we must represent what we wish to find. In every statement problem, you are asked to find some quantity. This time we are asked to find a number. So, the first thing we put down is Let n = a number Now we must represent the relationships given in the problem. One relationship is 5 more than twice the number. This can be represented as . Another relationship is three less than three times the number. This can be represented by . Now we must write an equation which used the relationship given between these tow quantities. What are you told is true of the quantities 2n+5 and 3n-3 in the problem? Write an equation using this relationship. Solve this equation to obtain the required number. The procedure outlined in frames 40 — 42 is the procedure used in solving all statement problems. First, represent the quantity or quantities you are asked to find. Second, represent any relationships which are given in the - 188 - 43. represent the quantity you wish 44. to find or represent a number as that is what you are told to find. Represent the relationships given in the problem. five times a number is 5p. twice a number is 2p. 45. problem or which you can deduce from the information given in the problem. Some- times there is a formula which gives the relationship between certain quantities. Make use of such a formula if there is one. Third, write an equation expressing a relationship which exists between quantities you have represented. REMEMBER an equation is an equali tye Fourth, solve the equation to get the required value or values. In order to do steps 2 and 3, you must read the problem carefully. You must also learn to interpret your algebraic expressions in words to see if they express the same relationship as the one given in the statement of the problem. Suppose we were to set up the equation and solve the following. Five times a number is 10 more than twice that number increased by 3. Find the number. First, you would . Let p = a number. (You can use any letter you choose. To see if you have the same quantities as those given, substitute your letter for p.) Next you would . Do this. Now we are ready to write an equation which will express an equality between these quantities or which involves these quantities. This relationship will be given in the statement of the problem or can be deduced from the con. ditions of the problem. -189- We are told that the number 5p is more than the number 2p increased by 3. So, comparing 5p and 2p+3, which represents the larger number? 45. 5p 46. Then to express an equality, what must we do with the 10? Be specific. Write the equation which expresses the given relationship. 46. You can't say add. You must 47. Remember you are comparing the say what you added to what. numbers 5p and 2p+3. You are Ybu can add 10 to the number going to complete the following 2p+3 because 2p+3 represents the so that you have an equality smaller number and adding 10 to expressing the relationship it will make it equal to the given in the problem. larger number. equation would be 5p = 2p+3+10. 5p 2p+3 or you could subtract 10 from 5p. equation then: 5p-lO = 2p+3. Since 5p represents the larger number, you could subtract some- thing from it to get a smaller quantity. Or since 2p+3 represents the smaller number, we could add something to it so it would equal the larger quantity. Solve the equation to find the required number. 47. The number is .2. 48. Set up the equation you could 3 use to solve the following. Find a number such that twice the number is 9 more than 1 1/3 times the number. 48. Your first statement must tell 49. The last equation of Zn -.E.n = 9, what letter you are using and 3 what it represents. states that the difference of two For example, let n = a number. numbers is 9. This is certainly Ybu are comparing twice a number true as one number was given as or 2n with 1 1/3 times a number being 9 larger than the other which is represented as‘fl n number, so their difference must 3 be 9. orasffporasl%n. 3 Solve the equation and find the required number. Pick out the larger quantity and then decide how to get an equality. _ 190 - Any of the following equations express a correct relationship between the quantities which are compared in the problem. 2n ='%2'+ 9 or2n-%n=9. 2+ 2 -9 =«— n or n 3 .22 49. The number is 13 1/2 or 2 . 50. Let w = the first number W‘+ 7 = the second number 2(W’+ 7) = the third number w + (W?) + 2(w+7) = 57 51. 4w + 21 = 57 w = 9 All of these are w+7 = 16 the required 2(w+7) = 32 numbers. You were asked to find three numbers in this problem. 52. Be Specific here when you represent the quantities you are dealing with. 50. 51. 52. You are dealing with.ppq ages for each boy. State Specifically To check statement problems, you must go back to the given statement. You can't check in the equation as the equation wasn't given to you. The equation was something ypp decided on. Going back to the statement - if I find twice the number or 2°.32 2 and also find %»of the number or 3‘21 is the first result 9 3 2’ larger than the second result? If it is, then the problem checks. Set up the equation for: The sum of three numbers is 57. The second number is 7 more than the first number and the third number is twice the second number. Find the three numbers. The sum of the three numbers is represented by adding the three numbers. Solve the equation and be sure to answer the question asked in the problem. Set up the equation for the fOIIOWing 0 George is 7 years younger than Jim. Jim's age two years ago was twice George's age then. their present ages. Find 53. You may have a different equation than the one given in the answer to the last frame. This will be true if you represented different relationships. Your equation - 191 - what you mean. will be correct if you used the For example, let a = Jim's present relationships given in the age statement of the problem. IT IS WRONG to say let 2 = Jim. 2 represents something about Jim The following is another way of not the boy himself. representing the information in the last problem. Solution: let a = Jim's present age Solution: let y = George's present then a-7 = George's present age age then y+7 = Jim's present age Two years ago both boys were two years younger. So, y-2 = George's age two years ago a-2 = Jim's ago two years ago y+5 = Jim's age two years ago a-9 = George's age two years ago Jim's age 2 years ago = Jim's age 2 years ago = 2(George's age 2 years ago) 2(George's age 2 years ago) So, a-2 = 2(a-9) So, y+5 = 2(ye2) This equation is correct for the above representation of the information given in the statement. There are other correct possibil- ities. If you have something else, make sure that when it is interpreted in words, you have the relationship given in the problem. Solve your equation. Be sure to answer the question asked in the problem. 53. Jim is 16 now and George is 54. Set up the equations for each of 9 now. the following. (a) 72 is divided into two parts such that one part is 18 more than twice the other part. Find the two parts. (b) The sum of three consecutive integers is 144. Find the three numbers. (0) A man is twice as old as his son. Ten years ago, the man was four times as old as his son was then. Find their present ages. (d) Van had twice as many marbles as Keith. Van lost 4 of his to Keith. Then'Van had 26 less than three times as many as Keith. How many marbles did each boy have to begin? 54. 55. _ 192 - (a) let b = the larger part 55. then 72-b = the smaller part b-18 = 2(72—b) or b = 2(72-b)+18 or You could have used the following representation. let y = the smaller part then 2y+l8 = the larger part Since the two parts must total 72, y-+ (2y+18) = 72. (b) let n = the first integer n+1 = the next higher integer n+2 = the next higher integer n + (n+1) + (n+2) = 144 (c) let a = son's present age then 2a = man's present age a-lO = son's age 10 years ago 2a-lO = man's age 10 years ago 2a—lO = 4(a-lO) (d) let x = number of marbles Keith had to begin then 2x = number of marbles Van had to begin After the game, Keith had xfi4 marbles and Van had 2xa4 marbles. 2xa4 + 26 = 3(x+4) (a) larger part is 54 and the 56. smaller part is 18. (b) integers are 47, 48 and 49. (c) the man is 30 and the son is 15. (d) Van had 20 marbles at the beginning and Keith had 10. Finish solving each of the problems given in the preceding frame. Make sure you answer the question asked in each problem. Sometimes we deal with geometric figures and need to know their dimensions, area or perimeter. The common figures we deal with are the rectangle, square, triangle, circle and cube. Rectangle - a four-sided figure where the Opposite sides are equal and parallel. a. Perimeter = distance '4 around. Perimeter = 2a+2b Area = ab Square - a rectangle with all sides equal. S Perimeter = 45 Area = 52 Write the equations which express the following relationships. (a) One dimension of a rectangle is four times the other dimension. If the perimeter -193- is 34 feet, find the dimensions. (b) One dimension of a rectangle is one less than five times the other dimension. If the area is 52 sq. ft., find the dimensions. (c) Find the area of a square if the perimeter is 56 feet. 56. (a) let x the length of one 57. The equation obtained in part b side of the last frame is p23 a linear then.4x+2 = the length of the equation, and so you are not other side able to solve it at this time. 2x + 2(4x+2) = 34 You should be able to solve both (b) let y = length of one side of the other equations. then 5yel = length of other side . y(5yal) = 52 Triangle - a three—sided polygon. (0) let a = lenth of one side Right triangle - a triangle then 4a = perimeter having a right angle. 4a = 56 The area is found by squaring The perimeter of any figure is one side and can be found after the distance around the figure. a value for a is obtained. Perimeter of a triangle = a+b+c 0K. Area of triangle = C: l where b is )}L 2 bh the length ,6. of one side of the triangle and h is the length of the perpendicular to that side from the opposite vertex. Circle - the perimeter of a circle is called the circumference and C = 277r . Area of circle =77r2. r stands for the radius of the circle. diameter = 2r. (a) Find the circumference of a circle of radius 4". e (b) Find the area of a circle of radius 1 1/2 feet. (0) One side of a triangle is twice the other side and the third side is 6 less than the first side. If the perimeter is 38", find the dimensions. _ 194 - 57. (a) C = 811in. 58. A rectangular solid called a (b = parallelpiped is a solid where ) A ‘211Sq' ft' all of the faces are rectangles. (c) let p = length of first side Parallelpiped then 2p = length of second side Volume = lwh _ and p-6 = length of third side ' (1 represents the length) p + 2p + (p—6) = 38 1 Surface Area = sides are 11”, 22", and 5". — _ K at the sum of the ur- areas of the faces and equals 2wh+21w+21h. A parallelpiped whose faces are squares is called a cube. | What is the volume I of a cube? 2_ ‘What is the surface \ area of a cube? 9.. 58. Volume of a cube = e3 where e 59. (a) Find the volume of a represents the length of one rectangular box which is 6" long, edge. 3" wide and 2" high. Surface area of a cube is 6e2 (b) If one edge of a cube is (2x-1) as there are 6 faces to a cube units long, represent the volume and each face is a square. of the cube. (c) The length of a rectangular box is twice the width and the height is half of the width. Represent the surface area of the box if the box has a cover. (d) Represent the surface area of the box given in part c if the box is open. 59. (a) V = 36 cu. in. 60. Set up the equations you would (b) V = (2x-l)3 = 8x3-12x2+6x—l use to solve the following. (c) let W'= width (a) A man wishes to enclose a then 2W’= length field by using 740 yds. of fence. and,l,w.= height If he wants the field to be 10 2 w yds. longer than twice the width, V = W(2W)C§) = w3 what should be the dimensions of the field? Surface Area = 2(g)(w)+2(w)(2w) (b)A salesman gets a salary of +2(y)(2w) $22 plus a commission of 45% on 2 all sales over $50. If he earned Surface Area = w2+2w2+4w2=7w2. $89.50, what was the amount of (d) without a cover, the surface his sales? area = 2‘(_w_) (w) + (2w) (w) +2(p)(2w) (c) The length of a rectangle is . 2 2 6 less than three times the width. or includes the area of only If the length were decreased by 1, five faces. the perimeter would be increased surface Area = 5w2 by 2. Find the dimensions of the original rectangle. 60. 61. -195- (a) let w = width 61. then 10+2w = length Perimeter = 2(w)+2(10+2w) So, 2(w)+2(10+2w) = 740 (b) let q = total amount of sales He received commission on all but $50 or on q-5O dollars. commission is .45(qe50) dollars. 509 22+.45(q-50) = 89.50 The above equation expresses the various amounts of money in dollars. You could have ex. pressed the amounts of money in cents. The equation then would be 2200 + 45(q-50) = 8950. Note that both sides of the first equation were multiplied by 100 to get the second equation and so the two equations are equivalent. Why? See frame 7 if you don't know why. (c) Let w = width of the original rectangle Then 3wa6 = length of the original rectangle wal = width of new rectangle 3wa4 = length of new rectangle perimeter of original rectangle is 2(w) + 2(3wa6). perimeter of new rectangle is 2(W-l) + 2(3W-L") e The periemter of the original rectangle is increased when the dimensions are changed, so the perimeter of the original rec- tangle plus the increase equals the perimeter of the new rectangle. 2(w)+2(3wa6) + 2 = 2(wel)+2(3wa4) (a) Let x = the smaller number 62. Then 6x-2 = the larger number 6x-2 _ 74% (d) Linen sells for twice as much a yard as rayon. For $30.60, I can buy 5 yards of linen and 7 yards of rayon. How much does each cost per yard? Solve the following. (a) One number is 2 less than six times another. ‘When the larger number is divided by the smaller one, a quotient of 5 and a remainder of 3 is obtained. numbers. Find the two (b) Evelyn bought a dress, a suit and a pair of shoes. The suit cost two and a half times as much as the dress and the shoes cost threeefifths as much as the dress. How much did she spend for each if the total bill was $102.50? (0) Jim left Lansing at 7 am and drove 50 miles per hour. George left the same place at 9 am and drove at 60 miles per hour. How long would it take George to overtake Jim? (Hint: what must be true when George overtakes Jim? (d) A boy started walking home at the same time his brother started out from home on his bicycle to find him. The boy on the bicycle traveled one mile per hour faster tham twice the speed of the boy walking. If they met after 2 hours and the boy was 10 miles from home originally, at what speeds did they travel? When the boys met, how far from.home were they? So far we have solved linear equations in one variable. For example, we have solved equations such as 3xa4(x+2) = 2x»?. 62. - 196 - The numbers are 5 and 28. Don't forget that the remainder in division is added to the quotient and is placed over the divisor. Emample: 13 divided by b = 6 1/2 or 6 + 1/2. (b) let 0 = cost of dress Then 2%6 = cost of suit and‘% c = cost of shoes 0+2 *2 a 2 c 5 c 102.50 dress cost $25, suit cost $62.50 and shoes cost $15. (c) The distance each man tra— veled must be equal to each other if one man overtakes the other. let t = number of hours Jim traveled then t.2 = number of hours George traveled Jim went 50t miles and George went 60(t—2) miles. So, 50t = 60(t-2). The men met after 12 hours. (d) let r = speed of boy walking Then 2r+l - rate of boy on bike 2r = distance traveled by boy walking 2(2r+l) = distance traveled by boy on bike Since the total distance is 10 miles, 2r + 2(2r+l) = 10. The boy walked at the rate of 1 1/3 miles per hour. The boy on the bike traveled at the rate of 3 2/3 miles per hour. They were 7 1/3 miles from.home when they met. x = 3 and y = 2 is a different 63. solution. There are an infinite number of solutions as you can name any two numbers whose sum is 5. We have also dealt with equations which contained more than one variable, and have solved these for one variable in terms of the others. The variable we solved for could only be to the first power so we were still solving linear equations. For example, axmbx = equation in a and a = c is a linear c-bx x providing x i 0. NOW’WG will consider linear equations in two variables. axmby+c = 0 is a linear equation in x and y if a, b and c are constants. The solution to a linear equation in two variables consists of a pair of values - one value for each variable. The solution of a linear equation in one variable consists of a single value. In the two examples given above, there is only one value of x which satisfies the given equation. Consider the equation xfiy = 5. A solution to this equation con- sists of one value for x and a corresponding value for y, such that the sum of these is 5. x = 2 and y = 3 is one solution Name one other solution to xfy = 5. HOW'many solutions to xwy = 5 are there? If we are given two linear equations in _t_w_q_ variables—(7m will refer to this situation as a gystem of equations) our prOblem is to find one pair of values - one for each variable - which will satisfy both equations. For example: Given the equations thy = 3 x-2y = 4 -197- 63. You can verify that x = 2 and 64. y'= -l is the common solution by substituting these values in both equations. 2(2) + -l = 4-1 = 3 so these values check in the first equation. 2 - 2(-l) = 2+2 = 4 so these values check in the second equation. Since these values check in both equations, they are the common solution. 64. x - 2y = 4, so x = 4 + 2y 65. 65. 2(M2y)+y = 3 66. 66. y = -1 67. to find the common solution, means we are looking for a pair of values - one for x and a corresponding one for y'- which satisfy both equations. Verify that x = 2 and y = -l is the common solution for these equations. The procedure of solving these equation is called finding the simultaneous solution or finding the common solution. Simultaneous solution by substitution. Solve one of the equations for one of the variables in terms of the other and substitute this into the other equation. é2x+y = 3 Choose one of the 4 equations and solve it for one of the variables. It really makes no difference which equation is used or which variable is solved for, but you are to solve for x in the second equation.' x-Zy Since this value of x must satisfy both equations in order to have a common solution, we may now substitute it in the other equation in place of x. What equation do you obtain when you do this? The equation we obtain will be a linear equation in one variable, which we can solve using previous methods. Solve the equation obtained in frame 65. This one value is‘pqp the common solution of the given pair of equations 0 What must you do now? What is the simultaneous or common solution of these equations? 67. 69. 70. 71. -198- You must find a value for x. 68. You will sometimes find the common solution written as (2,-l). common solution is x = 2 and This notation is referred to as y = .1. You must specify the an ordered pair. The first variable as well as the value. number of the pair is the value or x and the second one is the value of y. What values of x and y are represented by (-l,2)? x = -l and y = 2. 69. You can see that the ordered pair (2,-l) doesnot represent the same values as the ordered pair (-l,2). Find th common solution for 2x - 3y = 13 -3x+y =-9 we must first , then and then solve one equation for one 70. Do the indicated process. variable, then substitute this value in the other equation and Since there is no preference as then solve for both variables. to which variables you solve for now in which equation you use, choose the equation and variable where it is easiest to 501ve e In this problem, the easiest to solve would be in the equation. The easiest to solve is for y' 71. (29-3) is the common solution to in the second equation because these two equations, so in order the coefficient of y is l. to check this, what must be From the second equation, done? Y = -9+3x. substituting this in the first equation, we obtain Perform the check. 2x - 3(-9+BX) = 13. Solving, we get x = 2 and then solving for y we get y = -3. Common solution can be written as the ordered pair (2,-3). Be sure that you said substitute 72. Find t e common solution for these values in both equations. %X .E _ i If the values do not check in 3 y - 2 both equations, either the 6x - y = 3 solution is wrong or you have (Hint - remember you can.multiply made a mistake in the check. both sides of an equation by a constant getting an equivalent equation.) l... ..199- 73. In the last frame, we found an 72. Multiply both sides of the first equation by 6 to get the equivalent equation 3x+4y = 15. Now choose one equation and solve for one of the variables. The easiest to solve is for y in the second equation because its coefficient is -1. Doing this we get y'= 6xp3. Now substitute this in the other equation. 3x+4y = 15 becomes 3x+4(6xe3) = 15. Solve this equation for x and then determine the value for y. Solution is (1.3). Remember these values must check in both equations and this means in both opiginal equations. 73. (a) (8,-10) 74. (b) (-1,6) (c) (-3/7. 11/7) (d) (593) (9) (2/39 -4/3) equivalent equation for one of the given equations. Thus we f und the solution for §w= 15 6x- y'= 3 Since the above system is equiv- alent to the system given in frame 72, we can say that the the solution to both systems is the same. Solv and check. (a) 5x = -4y 2x-y=26 (b):2x - 4y = -26 5 X + 3y = 17 w. (c)(%x= 1 - .8y :\ x - y + 2 = O (d)<:jax+3y- 8a=O {y'+ 4a = ax Solve for x and y. eNm+sy=J k6x - 3y = 8 In any system.consisting of two equations in two variables, we can add or subtract the given equations and obtain another equation of the same system. Consider part e of the last frame. 7x+5y = -2 6xe3y = 8 Adding these two equations, we get l3xm2y'= 6. This represents another equation which will have the same common solution as the ones you solved. Check this fact by substituting the common solution If we had equations where the coefficients of one of the variables were the same, then when we added or subtracted the equations we would get an equation which contains only one variable. - 200 - For example, take the equations 2x-3y = 1* 4x+3y = 14. Adding these, we obtain the equation 6x = 18. This equation contains only one variable and can be solved for this variable using previous methods. In this example, the coefficients of y were such that the terms containing y were eliminated when we added the equations. we don't eliminate either variable when we add or subtract the equa- tions. We can change each equa- tion to an equivalent equation so that the coefficients of one variable will have the same absolute value in both equations. Change these two equations to equivalent equations such that the coefficients of x will be the same. 74. We obtain equivalent equations 75. Solving this system will give us by multiplying or dividing both the same solution as solving the sides of an equation by a non- given system. zero number. Now, since we wish to have a If we multiply both sides of the linear equation in only one var- first equation by 6 and both sides iable, what can we do with these of the second equation by 7, we two equations? will obtain equivalent equations where the coefficients of x in both equations are the same. We get 42x+30y = -12 42x-21y = 56 75. Subtract the equations. 76. Write the equation you obtain by this process and finish solving. 76. 51y = -68 77. Solve the following system by making the coefficients of y y = ~68/5l = -4/3 the same. Check. 3X-8y = -13 5011.1th!) is ( %,_ 3%) 2x..12y = - 7 77. To make the coefficients of y 78. In the problem given in the last alike, the smallest integer to frame, if you had multiplied both - 201 - multiply both sides of the first equation by is 3 and is 2 for the second equation. This gives coefficients of -24 for y in both equations. You can multiply both sides of the first equation by 12 and both sides of the second equation by 8, and then all the coefficients will be larger but in the same ratio as the ones iven below. 9X~24y = -39 xe24y = _14 Solution is (-5,-%). Remember that these values must check in both equations. 780 (3') (61’9) (b) ('393) (3) (2b, ‘b/Ba) (d) Exactly the same procedure is used here as in the above problems. Obtain an equiva- lent system of equations by multiplying both sides of each equation by some non- zero number so that the coefficients of one of the variables is the same in both equations. To make the coefficients of y the same, multiply both sides of the first equation by a and both sides of the second equation by b. Solution ( Zab 9.232323) a2+b a2+b2 (e) You will need to use the fact that when two factors are multiplied by -l, the value of the fraction is unchanged. 3 1 ti .:l:.;:l ° u °n ( c-d c-d) sides of the first equation by -3 and both sides of the second equation by 2, then you would have had to add the two equations in order to eliminate the terms containing y. Solve each of the following for x and y and check. (a) 2x+3y = -15 3X-5y = 63 (b) 4x+3y = 7 5x-éy = -4 min-3w = 5b 7xfi6ay = 12b (d) ax+by = b bx-ay = a (e) cx-dy = -l dxhcy = 1 79. We have dealt so far with two linear equations in two variables which have a common solution. Not all such equations have a common solution. Linear equations in two variables which have no common solution are called inconsistent equations. For example, if we tried to solve 3xv2y = 5 6x+4y = 2 we would find that no matter which method of solution we used, we would get an equation which con— tains no variable and which is of the form, 0 = some nonzero number. An equation of this form can never be true. Let us multiply both sides of the first equation by -2. We would get the equivalent equation -6xp4y = -10. Now if we add this to the second equation, we get 0 = -8. This equation is never true. Therefore, the equations have‘gg common solution, and are called equations. - 202 - 79. inconsistent 80. Sometimes we have two equations in two variables and they have an unlimited number of solutions. Equations which have an unlimited number of solutions are called dependent equations. When dependent equations are solved by either of the methods we know, we obtain an equation which is an identity. Identities are equations which are true. 80. always 81. Solve the system 4x-3y = 6ye8x+l8 = Identities are true for all by substitution. values of the variables. 81. In substitution, you solve one 82. Since we obtain an identity when of the equations for one of the we solve these equations and variables in terms of the other identities are true for all values variable and then substitute of the variable, we have a system into the other equation. of equations which has an unlim. Solving the first equation for x ited number of solutions and we get x =‘2E2y . a system of equations which has an unlimited number of solutions Substituting this in the other is known as a system. equation, we get 6y;8(2E21)+18 = O. This simplifies to O = O. 82. dependent 83. Solve the following sets of equations and state whether they have a common solution and so are consistent equations or whether they are inconsistent or dependent. (a) x*y+2 = l-6y = 6x (b) 3(2x-5)+5(3y+l)= (d) 3X-2yh7 = 42+12y = 18x - 203 - 83. (a) inconsistent (b) consistent, solution (2,0). 84. (c) consistent, solution (-l/3,l) (d) dependent (6) dependent (f) inconsistent 84. the numbers are 23 and 47. 85. let x = the first number let y - the second number Equations are x-9 = 2y 3y= x+6 85. 86. (6) 3X = 5-2y 4y = lO-6x (f) 5X-3y—7 = 0 le = 6y+2 Sometimes it is more convenient to solve word problems by using two variables and two equations than to use one variable and one equation. The procedure is exactly the same as when one variable is used. First, represent the quantities you are asked to find. Then analyze the problem, repre- senting the other relationships given and write two equalities if you use two variables. Then solve to find the common solution. To solve the following. The sum of two numbers is 70. One number is one more than twice the other. Find the two number. Solution: let x = one number let y = other number x+y = 70 expresses the relationship that the sum of the two numbers is 70. x = 2y+l expresses the relationship that one number is l more than twice the other. Find the common solution of these two equations. Set up the equations you would use to solve the following. Use 3E2 variables. The first number is 9 more than twice the second number. Three times the second number is 6 more than the first number. Find the two numbers. Your equations may be set up differently from these, but they must express the same relationships. Make sure than when the equations are interpreted in words, your equations express the same rela- 86. first number is 39 and the second number is 15. - 204 - 87. tionships as those given in the problem. Solve the equations you got in frame 85 and find the two numbers. Set up the equations you would use to solve each of the following. Use two variables. (a) If 12 yards of cotton cost the same as 5 yards of rayon and if 3 yards of rayon and 5 yards of cotton cost $6.10, what is the price per yard of each material? (b) The charges for a telegram consist of a flat rate for the first ten words and an additional charge for each additional word. If it costs $.98 to send a telegram of 17 words and $1.26 to send a telegram of 24 words, what is the flat rate and what is the cost of each additional word? (c) In a certain rectangle, the length is h less than three times the width. If the width is in- creased by 2 and the length is decreased by 3, then the perimeter is decreased by 2. Find the dimensions of the rectangle. (d) A merchant bought some radios. He got 8 of one kind and 4 of another kind for $128. The next time he ordered, he received 9 of the first kind and 3 of the second kind for $a more. What was the price of each kind of radio? (6) A man traveled 280 miles. If he had traveled 10 miles per hour faster, it would have taken him 40 minutes less time. What was his rate and time? (f) In a number such as 23, the 2 is called the tens digit and the 3 is called the units digit. To form the number, the tens digit is mul- tiplied by ten and the units digit is multiplied by l and the results are added. Thus if the tens digit 870 (a) Let a = - 205 - cost per yd. rayon 88. Let b = cost per yd. of cotton 12b = 5a 3a + 5b = 6.10 Both a and b represent the cost per yd. eXpressed in dollars. If your variables represent the cost in cents, your second equation will have 610 instead of 6.10. (b) let x = flat rate let y = cost of each additional word x + 7y = 98 when x and y are x +14y = 126 expressed in cents. (c) let w let 1 width length perimeter is 2w+21 new width is w+2 new length is 1-3 new perimeter is 2(w+2)+2(l-3) 2w+21—2 - 2(w+2)+2(1-3) l 3wa4 (d) let p = cost in dollars of each radio of lst kind cost in dollars of each radio of snd kind. 128 132 let q 8p+4q 9p+3q (9) let r = number of miles per hour at which man traveled let t = number of hours rt = 280 (r+10)(t- 2/3) = 280 (f) let t = tens digit and u = the units digit. number is 10t+u. of a number is 7 and the units digit is 4, the number is 7(10)+h or 74. Let t = the tens digit and u = the units digit and represent the number which this forms. Write the number where the hundreds digit = h, the tens digit = t and the units digit = u. Solve each of the preceding systems of equations. Solve each of the following problems. You may use either gone variable or two variables. (a) In a certain two digit number, the sum of the digits is 11. The number is 20 less than twice the number which is formed by rever- sing the digits. Find the number. Hint - when the digits are reversed, the tens digit becomes the units digit and the units digit becomes the tens digit. Thus the number 24 becomes the number #2. (b) Sue's present age is one year more than 3 times her age five years ago. Her age in five years will be twice what it was last year. What is her present age? (c) A boy started for a place 23 miles away. He started by walking and walked at the rate of 1% miles per hour. Later he was offered a ride and completed his journey at the rate of 40 miles per hour. If it took him 2% hours to get to his destination, how far did he walk? - 206 — let h = hundreds digit let t = tens digit let u = units digit number is 100h+10t+u 88. Solutions to systems of equations in frame 87. (a) rayon cost $1.20 a yd. and cotton cost $.50 a yd. (b) the flat rate was 70¢ and the Eost of each additional word was ¢. (c) length is 17 and width is 7. (d) the first kind cost $12 each and the second kind cost $8 each. (e) In solving these two equations, first do the multiplication in the second one. Then subtract the two equations. This will give you another equation which belongs to the system and when this is solved with one of the original ones, you will get the required solution. This equation is 101; ... -2-r __ 29 = 0. Solve this with rt = 280 using substitution. Remember the object is to get one equation in one variable. However, when you do this, you do ‘ggt get a linear equation so you cannot complete the solution at this time. The solutions for the three additional problems in frame 88 are as follows. (a) let t = tens digit let u = units digit t + u = 11 lOt + u + 20 = 2(10u +t) the number is 74. (b) let a = Sue's present age a-5 = her age 5 yrs. ago a+5 = her age in 5 yrs. a - 1 = 3(a-5) a + 5 = 2(a-l) Sue's present age is 7 yrs. (c) let h = number of hours walked then 2%: .. h = number of hours rode l%h+40(2-21~-h) :23 He walked 3 miles. Note - h isnot what is asked for. negative exponents. Chapter 7 - Exponents and Radicals In this chapter, we shall define the meaning of fractional and 1. (a) m5n5p6 (b) 36a12bl5 (e) xiv (d) xllpzy Be careful here. An 1. -207... We will also define complex numbers. First, we will review the definitions and laws with positive integral exponents. If n is a positive integer, then an means a°a°a- ... for n factors. If m and n are positive integers, then am ~ an = am+n. If m and n are positive integers and m>n and a 74 0, then am - mpn ‘a'- a . a If m and n are positive integers, then (am)n = am“. Simplify the following. (a) (m3nnp)(m2np5) = (b) (32anb5)3 = «we: .2. (a) (x3p2)“(x2y) = (xp2)3 There are times when we need to know the meaning of an exponent of 0, and also the meaning of a negative exponent. We shall use the following definitions. a0 = 1 providing a # 0. a'n = -l- providing a # 0. an (a) x0 = , (b) 12° = .. _._: (d) (2P)0 ______9 9 (f) 2-3 0 (e) 2p0 (e) 2‘1 Express the following with positive exponents and combine. (a) XZyo - 208 - exponent affects only the letter, (b) 3x-2y3 number or symbol which immed- iately precedes it, so the 0 (c) (2x'3)0(2‘1x3) af ects only the p. 2p = 21 ~ p0 = 2'1 = 2. (d) 3x“ +(3x)'2 (e) p'1+q‘1 (r) ha to to JP \n n I NIH numbers and fractions in this way, we can use the same laws as those given in frame 1 for positive integral exponents. Thus, al/2.a1/4= al/2+1/4= a3/4. a3/5 81710 = (al/4)2 = 7. a3/5 ‘ l/lO = a5/10 = a1/2 8. Since we have defined the meaning of negative exponents, we can now aCL/zl')2 = aZ/u = a1/2 define division in the following way. 32,: am‘n when.m,# n n a0 when m.= n When m(n, then mm is a negative number, but by using the definition of negative exponents, we can write an equivalent quantity having positive exponents. Simplify the following. Write the final answers with positive exponents. (a) 251/2x2/3y...2 . loo-l/le/zy (b) (asp-“q5/3r3)1/2 (c) 6’+"la2b3/’+c”3 4172a'6bc"3 (d) x3221-2 (Xty) ‘2 8. (a) You can multiply by adding 9. In the last frame, we had x2/3. We exponents only if the bases are have defined the same. The numerical bases in Xl/n.aé\Q/§ providing the nth this case are not the same. .5x Z/By -2 . l ;l72 lgx7/6 -1 root of x exists. The nth root of /6 10K /§ 2 y x doesn't exist at this time, if 27 (l/y) x7 the nth root isnot a real number. 2y - 210 - (b) 361/2(p"l+)1/2(q5/3)1/2(r3)1/2 For example,\/:4 isnot a real number because we can find pg.real = 6p"2q5/6r3/2 = 695/6r3/2 number such that when it is 2 multiplied by itself, we get -4. P (c) When the negative exponents Is‘2/-8 a real number? are changed to positive on , the numerator becomes a2b3?fi . Is QJ-lg a real number? 64c The denominator becomes _22_ . a5 03 Dividing the numerator by the denominator, we get a7 . 128b1 (d) The numerator becomes —l+—l= ° The x2 y2 x2y2 l denominator becomes . m Dividing, we get (3:1)2(23+x2) , x3y2 9. -8 = —2 so yes it is a real 10. xl/n or n x exists when x is a number. real number. n can be an integer greater than 0. There is no real numbe such This also exists when x is a that a‘a'a'a = -16 so‘éj-l6 is negative number and n is aniggd get a real number. integer. xp/q will be a real number when or when . 10. x is positive and p and q are 11. xp/q = (xl/q)p = (x.p)]'/q = Q/xp = any real numbers except q = 0 or CQ/§)p providing the qth root of when x is negative and q is an x and the qth root of xP are real integer. numbers. 2 l 2 27 /3 - (27 /3) - = = . Express the final answer without exponents. ll. (272)1/3 = 32 = 9. 12. Evaluate each of the following. You can see that it is easier (a) 82/3 (b) 85/3 (c) (--8)5/3 to find the cube root of 27 and _3/2 _3/5 _3/5 then square the result than to (d) 4 (e) 32 (f) (-32) square 27 and then take the cube 2/3 2/3 _3/2 root of the result. However, (g) (~125) (h) -125 (i) -16 both procedures are correct. -4/3 <') 163/“ (k) (-27) J (1) -27'4/3 _ 211 - 12. (a) (\1/8)2 = 4 13. Note carefully the differences between parts g and h. Also note (b) (\Q/—)5 = 32 the differences between parts k \ and 1. If you made errors in the (c) (~3/-8)5 = (-2)5 = -32 last frame, it would be advisable to write out the meanings of these (d) =l. in terms of the radicals. Remember (\f4)3 8 an exponent affects only the letter, number or symbol which immediately precedes it. Thus, in part h, only 125 is raised to the power, and (e) (in-3 = $31 in part g, -125 is raised to the (r) 02/732)-3 = -Lg. power. (g) (‘2/-125)2 = (.5)2 = 25 As stated before, the laws of 2 exponents as given in frame 1, (h) - ( Q/lZS) = - 25 apply whether the exponents are integers or fractions and regardless (i) -(\(16)'3= -(4)”3= - of whether they are positive or negative. (3) (\fi/16)3 = (2)3 = 8 Simplify each of the following. . Express with positive exponents. (k) 03/357)”: (-3)"4 = _1. 81 (a) (3al/2b'3c2)(4a3/2b'5c‘3) <1) -<\3/2’7>"‘= -<3>-‘* = - .1— 81 (b) (4ab'zc3)1/2 (c) (incl/21213 -10X3y-2 (d) (-p2q‘3r5)'5/3 (e) EZapb'3q13 8's: -212. "Zpbq (f) (81a _)-l/2( b2 w) (b"’1(27ab"2 13. You can remove the negative 14. If we are to express (x"’2 +y)'l exponents first and then do the with positive exponents, it is operations or you can do the usually to our advantage to remove Operations and then remove the the negative exponents first. negative exponents when the However, be sure to remove only Operations are multiplication, one negative exponent at a time. division and raising to powers. 2 8 l. 1 1 X-2”)-l= -—-l = l (a) 12a b' c” — 12a2--3.' - -2 1 be C X ‘3’ $4.37 _ 12a2 bgc This is a complex fraction and to (b) 41/2H1/2b-l 3/2_ Simplify it, we must . 2:1/2o 3/2 Do the simplification. b (c) 125x3/216 = 8 25y -lOXBy'2 2x3 2 (d) (_ 11%5/3p'10/3q5r-25/3 = 10,3 q25/3 P r (e)-8a3Pb-9q = 2.53... -16a'2qu 21310q (f) 81‘1/233 ° b‘?_ 3 2 a ..b-flé— 27'132 9b5 327?: 14. Multiply both numerator and denominator by the same number namely or express the denominator as one fraction and divide. 1+x2y 15. (a) x-‘Ly-Lt = its“ X y (b) x—l/2_ye2/3= 1_y2/3x1/2 1/2 X (C) x"6-27y"6 = 6-2 x6 X (01) 8x‘2+v'1= moi xzy (e) x"3_y-3= 3:3...)(3 x3y3 — 212 - 15. 16. You must note that (362437)"l doesnot equal xzfy‘l. You cannot multiply the exponents to raise to powers unless the the base is composed of factors. Perform the indicated multiplications and then express the products with positive exponents. (a) (X'Zty'2)(X'2-y'2) (b) (x-1/4w1/3xx-1/kyl/3) (e) (X‘Z-By 2)(X' 4+3x72 y'2+9y'4) (d) (2x ‘2/3 wl/3)(4xé“/3- 2x“ ~2/3y ~1/3 +y 2(3) (e) (x73/2-y’3/2)(xf3/2+y’3/2) You will note that in parts a, b and e of the last frame you were multiplying two quantities - one of which was the sum of two numbers and the other was the difference of the same two numbers. When two quantities of this sort are multiplied together, the result is the difference of the squares of eagh of the terms. Thus, in part a,x is the square of 2x" and y'4 is the square of y"'2 . Consider parts c and d. What relation do the terms in the product have to the terms in the binomial which is given as one of the factors? 16. l7. 18. 20. 21. They are the cubes of these terms. - 213 - In part c,x 6 is the cube of of x‘ ‘2 and y"6 is the cube of y‘Z. l7. (xi/1W3)(,C2/3h4/3y1/3gi/3) 18. (xi/Z-yl/2> the square roots of negative numbers are not real numbers. No because cube roots of both positive and negative numbers are real numbers. (a) as squares (ESE/€335347d)fi (x7 2+8y’2)(X":- 8y’2) l3+ZX-4/3y~4/3 +l6y“ 8/3) :b)3?§ -sq1>f:§3/22+y-1> glub-Z/B)(X-2+X-ly~2/3+y-4/3) ho 3as gquar s 3"'3’ )(x as cubes (XSZ ~y2)(xA v-3) +x‘2y2+y”) 19. 20. 21. 22. Then we can considerx 6 -27y' '6a the difference of two cubes andas factor it as we would the difference of any two cubes. Consider x-y as the difference of two cubes and write its factors. We can also consider x-y as the difference of two squares if x and y are positive numbers. Consider x—y as the difference of two squares and write its factors. Why do x and y have to be positive numbers in order to write the factors of qu considering this as the difference of two squares? In factoring xey as the difference of two cubes, do we have to limit the values of x and y? Give a reason for your answer. Factor each of the following in two ways. First consider each as the difference of two squares and second consider each as the difference of two cubes. Donot change to positive exponents. (a) x'4-64y'u (b) X'B-y'z (c) y“ / Remember the sum of two cubes can be factored but that the sum of two squares cannot be factored. You should also look for common factors and try to factor trinomials of the quadratic type. Factor each of the following. The numbers in each case give a clue as to hogg gactor each one. (a) 27x fig (b)x (c) x:5:16x"3 Hint: remove common factor first so that exponents in one faftor are positive. (d) 3x 2 -13x '.30 22. 23. 24. - 214 - 230 (a) as cubes (3x‘ +2y-1><9x- +6x-2y*1+4y-2> (b) (X’l+2y'3/2)(x’l 2y'3/2> when considered as squares. (0 remove a co on factor of x- . (1-16 ) = x” (l-4X)(l+#x) (d) (3X +5)(X'1-6) 5 .3y -1 -2; -3X +9y y2-3x3y+w 672 X y ux—Lt-lsy-Z: L" 2. 2X4 x (c) x'5 _ x2 x'7(l-2x) l—Zx (d) 1 X172 172 -y .-L+/3 W-2/3y2/3 W3: 1).:X2/3y2/31/3x5/3 x#73 (a) x'6-° 2#. (b) (e) (a)x = 2 25. (b) Z-x = 5, x = -3. (c) 2x = x-Z, x = —2. (d) Here we don't have an equgtion of t e form am=an , bt 22 so x =32 becomes = and x=2. (e) Write 27 as 33. If x3=33, then x = 3. (f) 3x-2 = 2(x—l), so x = 0. Reduce the following fractions to lowest terms by first factoring. Express final results with positive mats. _:_;§§ZZ__ X-3+3y-lu (b) 16x 8 -25y uxfu+5yf 2 (c) :55+2: '4 We can solve equations involving exponents if they are of the form am = an. In this case, we have an equality involving exponentials and since the bases are the same, the exponents 2Igust be equal. Thus if 2x = , x must equal 5. Also, if the exponents are equal, the gases must be equal. Thus, if x = 23, than x equals 2. In other words, exponentials are equal if the bases are the same and the exponents are equal. Solve for x in each 05 the gollowing. (a) x3 = 23 (b) 3 "x; (c ) 42X = ”x-2 (d) x5 = 32 (e) x3 = (f) 53X-2=52(x-l) If equations are not given in the formam = an, then we must put them in this form beforg solving. For example, to solve 3 ‘BL - 27, we must use the laws of exponents on the left side of the equation and express the right side of the equation as a power of three so that we have the same base on both sides of the equation. 32 . 3X = . 27 = . _ 215 _ 2 25. 3 ° 3X = 32+x 26. 27 = 33 26. 2+x = 3, x = l. 27. 27. 28. 4 has to be expresged as a power of 2 or as 2. On the right side of the equation we have to raise to a power and obtain 22x .4 The equation now reads 22:22 x— So 2 = prh, and x = 3. 28. (a) 2""1+3x = 23 29. x~l+3x = 3, x = l (b) 512X~53= 52 12x+3 = 2, x = -1/12 (c) (22)2-23+x=2“, 2u°23+x=2“, 4+3+x=4, x = -3. (d) 33.3 2X: (3,2)2X-1 33+2X= BhX-Z 3+2x = #x-Z, x = 5/2 30. 1/2 2/3 29. $22112 +S2X~22 _ (2X- 3)1 3 ( l 2 - 3y+l) The LCD is (2x.3)1/3(3y+1)1/2 (3y+l)l/2(3y+l)l/2+(2x-3)2/31/3 (Zx-B) (2x-3)1/3(3y-+1)1/2 30. (3y+1)1 or 3y+l 31. 31. (2x-3)l or 2x-3 32. 3 Now we can write 32+x=3 . Solve for x. Solve for x: 4 = (2x-2)2. Solve the following. (a) prl.23x= 8 (b) (53x)“°53= 25 ( ) 422 3+x_16 (d) 27° 32x =92“l To simplify the following, we would first express with positive expo- nents and then.combine. Simp 11 (3y+1)f72(2x_ 3)‘1/3+(3y+1)‘1/2(2x 3)2/3 Let us consider the first term in the numerator of the answer to the last frame to see if it can be simplEfied f ther. (3y+l) (3y+l)1 2 is a situation where we can apply the law of multiplication for exponents. This law states thata m.an=am+n or that the product of two expo— nentials which have the same base is that base to the sum of the powers. So. (3y+1)1/2<3y+1)1/2= . <2x-3)2/3<2x—3)1/3= ___. Then the fraction given as the answer in frame 29 can be rewritten as an equivalent fraction which involves no fractional exponents iggfilwegiw (3Y*l) 2(2x_3)1 3 - 216 - .;22:l+2x-3 33. Simplify each of the following so 2(3y+1)172(2x _3)]-/3= that the answers are expressed as a single fraction and with positive 3y+2x-2 exponents. 1/41/2 3/4 (3y+l)1/2(2x-3)I7§ (a) (SK-j) (x::)12+(5x-2))-1/2 3/5 (b) <2y+7>1 3<3ya1> 5+(2y+7>-2/3(3y-1>3 (c) 31/3 (e) 2x2(X-2y)—2/5-(x~2y)3/5 33. (a) jx+4)1/2 (5x .2)?/4 = 34. Ne are now going to consider the (5x2)11/E (r+u)l_/2 radical form instead of using X- fractional exponents. Sometimes 6X+2 it is more convenient to use the (5x-2)1/hkv+4)1/E radical form than the exponential A form. (b) 613+19y-6 We defined al/n to be the nth root 2/3 375 of a providing this was a real .+ - (2y 2) <3J”1) number. 1/ n (c) 3<2x +3)-(x2-1)_ _gx :+10 That is, a‘ “:\\/E . (x2_1)17§ (X2 _1)1 2 We shall need some other laws con- (d) 2y—3(3Y-4) : 12-7y cerning radicals. These laws can (3y—4)2/3 (3y— ”)2/3 be derived from the laws of exponents. (e) ZXZ-(ngy) = 2};2-X+gy ‘Q/a“ = a because \0/23 = (an)l/n = 2 5 275 .-2 _ (I y) (x 2y) an/n = a1 = a. \Q/Zb = (ab)1/n = al/nbl/n. 3-4 = 3“» Q/ya = Y 35. We can apply these two laws to l u reduce radicals. V7713- __, 3 / ,ul/u ’ For example, W zw _-_- \2/537- = \Q/Eg °\Q/§'= 2\Q/§ . \VT' _ 35. V2 23 =2V 223 =\(f2_ V3 = 2v- 36. Reduce the fol§7%. (C)\/3_2_ (a) 75 (b) (e) 300 (f) 1082 ngééggg (h) 1 2 (i) (j) 98 (k) 32 (I) 1728 -217... 36. (a) . -..-. Q5 37. The same principles can be applied (b) 2 '2 = 2 2 (C)\QjEETE = 2‘k/5 in reducing radicals such as a7b . (am = 10V? figfia Vivi? . b = a°a°a°bfa V6 (r) Viz—3773 = 2-3V'5 = 6V3 Reduce the following. \2/337§ = s\%/§ (a)\/Z335 (b)\/;ii§5 (h) 5 25-6 = 2% MW WW \QyEF:E = 3\&/§ (e)\%/S3§3 (r)\J5§;§EZB \/?§TZ = 7\/§ \2/552§¢;§ \€/;IU§I2;8 (aw-273775 = 1275 (MW = 2-2-3 = 12 37. (a)\/;§'a2°a°b2'b2°b2 = 32h3V§ 38. If we are to reduce\/;?5+t+xy-my~2 W - then we must use the same laws (b) \/::<:2°x2wtz‘x‘g’xz'x'y‘goyzoy'Z'y‘d as we used in the last two frames. = xjyu'v; We'lcnow that ‘Q/a“ = a, and in 3 _ this case we are looking for a (c)\/a3°a'b3ob3 = abzwa square root. Thus, we wantVaTz so that we can apply the law. \2/g3-x3-x3.x3.y3.y3.y2.23.z 4 We also know that Q/ab =\Q/a\th->. = x yzz £5122 4 Can we apply these laws innnedi- (e) \[p3°qioq2 = qW ately to smplifyvm ? (f) V42.3.x,y2.y2.z2.22.z = Give a reason for your answer. 43’222 3x2 (g) W5°6°ry3°rzj°zz = ZyZW (h)\€/XLU’YLU‘yZ‘z5‘Z3 = x2y22\é/§ZE3 38. No. We have to have factors in 39. What must be done first to order to apply A simplinyx‘fl-ny-I-hyl ? 39. factor the polynomial 1+0. Simplify VXZ‘WWYZ . 40. \Figy) (x-pzfl 2W #1. Reduce \[(x£-9Y2) (DC-3377 . "' y - 218 - al.\/(x-3y)(x+3y)(x-3y) = 42. \((x-3y)4(x+3y) = (ac-3y) Vx+3y 42. (x-By)Vx*3 = x‘Vx+3y -3y‘¢x+3y 43. or the quantity x-3y is multiplied by the square root. In x-3y\/# nly 3y is multiplied by the square root. 43. (a)\/in—75z = 2x»? (b)K/Z3x-5525¢(2x+152 = (BX-5)(2x+l) (c) Vx2(x.l+)-le(§3Jy = \/(x-4)(xZ-1<§ Axe-meal (M = (x-uNx‘J 111+. H5. 44.3[18 913302 __. $12 vs 5 5 Note carefully the answer to the last frame. (x-By) Vx+3y is _n_9_t_ the same as X'By VJC+3y o What is the difference in these two quantities? Simplify. (a) VExZ-ZBM9 (b) V(3x-5)l(2x+l)2 (c) vxmxéléfitfi To deal with fractions which occur under the radical sign, we need one further law. a l/n 1 n 0-) = ii7; By applying this lewd/3 :\[-L_ V 4 \/E A5. , 2 Simplify:\vfiE§ 25 ’ Simplify each of the following. (a)\/§_% (b) 2:; b y (C) 3 125b _2x4+30x+25 x2-lux+49 (e)\//(x-3)(x‘- -6x£§7 2503:3315 211 (f) 3 8-221 27y3 (g) 2x+5 (d) er 2x+5 45. (a) y2§2 =VZE'E E- 6Y2 LIE. In 2x«Ear art g of the last frame, we got V206 V775 7V‘2’ and this equals 1 as it -219- (b) 6Z¥aZ¥a°b40b432 = 6abfy2a x °y °y2° xy3 (c) 933-2a3a2 = 2a gIZaZ \2/534: 5% (d) 3163:4552. = 3.2222 \i (x-7)2 x"? (e)\?Kx-J)T_x_-3) (xaL 53t2x 2g92. 2 Siy (f) 8.2 3 8—27y can be factored, but it is not a perfect cube. It is the— difference of two cubes. \le.2)‘*(2x+5)2c2x+5l \/2 5 __. Lac-zlzlgxfi) V2X+5 = (x—2)2(2x+5) via-5 #6. The square root of a negative 47. number is not a real number. represents a number divided by itself and a number divided by itself equals one unless the number is 0. However, in this case, we need the added condition that 2x+5 15.222 a negative number. Why can't 2x+5 be a negative number? The square root of a number which is not a perfect square is an irrational number. The cube root of a number which is not a perfect cube is an irrational number. The general case is the nth root of a number which is not a perfect nth power or, in other words, the nth root of a number which cannot be factored into n identical factors is an irrational number. At the beginning of the term, we defined a rational number as one of the form‘a where a and b are b integers and b i O. The definition of an irrational number then is . 47. numbers which can't be 48. \/§, «/—, Q/Z, ‘¢3 72 are examples expressed as the quotient of numbers. two integers. 48. irrational 49. Is Y1? an irrational number? Why? 49. No. -# is not a perfect square 50. so it isn't a rational number. If you are in doubt, refer back to frame 9. - 220 - In addition, the number under Irrational numbers of the form the radical sign, must be such ‘a can be written as‘l that a real number taken as a b b a ' factor the required number of This is called the rationalized times equals the number under form of an irrational number. the radical. Thus,\/:E = a only if a’a = -4 where a is a real \f; = ‘aE because a/b and a'b/b2 number would makevj; a real b b2 number. Since there is no real are e uivalent fractions. number which when multiplied by Q _Qafi _@ ._. l V313 itself equals .14, VI is _r£>_t_ an b2 — A?— b b . irrational number. \ Note .. irrational numbers are Expressv'j in rationalized form. real numbers. 2 50. 2 = = % or %Y5 51. Express 2 in rationalized form. 2 E 2 8 5LV§ =V55 =m =YZIQ 52. The object in rationalizing is to 671 W 8 write the number under the radical as an integer and this means that the denominator must be a perfect square for it to be brought outside the square root sign. In the last problem, instead of changing g to O we could have '61? changed it to 10 In both cases '13 o the denominators are perfect squares and so we can find an equivalent fraction where we have a positive integer under the radical. If both of these procedures are correct then we should have equivalent answers. How would you determine if 10 and SZEOare equivalent fractions? 8 52. reduce the fractionfi 53. Reduce \jgo 8 O or multiply both numerator and denominator of? by the same number to see if it equals\/Ea/8. 53. $.10 z zaylo =\zlo 54. Thus, you can see thatVE can be 8 56. 3 12 2-2'~'2' e L12 3252. - 221 - 55. 56. 57. (d rationalized by changing either to or tom}; . It is usually to 8 your advantage to change to the smallest equivalent fraction, which has a denominator which is a perfect square under the radical. Rationalize. .4; 27 Express your final answer in lowest terms. Either procedure given in the last answer is correct but you will no- tice it is quicker to use the second one. Use the Erocedure you understand the best. Rationalize" __. and express your 75 final answer in lowest terms. You will notice that in this last problem, both numerator and denom- inator of the fraction under the radical were multiplied by 3. In order to determine what to multiply by, factor the denominator into a factor which represents a perfect power and another factor. Multiply both numerator and denom. inator by a number which will make this second factor a perfect power. You will notice that then the two factors do not need to be multiplied together. It is not wrong if you do the multiplication, but it is not necessary.‘ To rationalize 3 3 you can use the same procedure, but this time the denominator must be a perfect cube in order for a positive integer to be left under the radical. Rationalize\3/§ 2 O Rationalize each of the follo g, (c) a (a»vl:Zi (b)\a/— 28 b ' (983140; 63a2b5 27 ng. - 222 - 57. (a) 21 =\[2l-§ = 2312 = 58. If we are to rationalizeyi "9°30 V9735 we can follow either of two grocedures. We know that V33 = WW so we can multiply both numerator and (b) 3 10_3[10 denominator by a number which will give a perfect power in the denom— (c) = a ab inator under the radical. 2‘9 is M . 7 3V" = = (d) 28 73£vb=llbq lib: v; vgvg ‘I9.7.7azb5 3°7ab3 252a Eb Bab3 (e) x $41le 3 58. V15 __. E 59. Instead of this last procedure, we V37; 5 know that(_a_)l/n ._._ al/n and so b i175 \E = 2 and this can be rationalized v3 5 as we did previously. Rationalize the following. (a) V5 (ML 8 (0)351 V7 V32 (d)‘ /20a3 (e) 3 (r))[8x5 7b x-3 m 5°. (ME (bvgggfl = l 60. What answers to the last frame (c2 :2 2 represent irrational numbers? (d 14’ ° a3b :jiabié W513“ h (x 3)2 x (f) V4.21; /x-y =2x2 VE-Ux-y \[x-Ty- x-y x-y 60. all except b 61. We can say that Ya; V33 = Vii-5' becau e we (ab)1y7WY/norw=%% . an we use this law to multiply byV-S- ? Give a reason for your answer. _223- =l-Pl/3 andV—= 62. M7381nce we have different root? ‘we cl éyse the law (ab)? n as in this law all quantities are raised to the same power. 61. No 62. index 63.\QIE 63. (42)l/6(53)l/6= (42.53)l/6 64. 6b. (a) 21/3.21/4= 24/12.23/12= 65, (24.23)1/12 = 27/12 =W So in order to multiply irrat- ional numbers, we must have the same root. The number which indicates the root is called the index number. InWE, the 3 is known as the number. = 41/3 because of the definition of fractional exponents. 41/3 = 02/6 because 1/3 and 2/6 are equivalent fractions. \[5 = 51/2 and this equals 53/6. so\2/E’o\/' e 41/3.51/2=n2/o,53/o. Note that in the last quantity, the roots are the same in both cases, namely the roots are 6. Now using the meaning of fractional exPOnentS. 4.2/6.53/6 = ( , )1/6 (#2 53)1/6=(2000)1/6 This can't be reduced as 2000 is the product of three 5's and two 4's which is the product of three 5's and four 2's and there aren't six identical factors here. Sometimes it is possible to simplify radicals by changiflg the order. For example, can be written as (32)1/4 which equals 31/2. When the order is changed, that is when the index is changed, note that the base is also changed. (a) = (b)\/- 27 = = = ()‘Q/i'e’=_— = = <§>\2/W—12=__'7=Z_=_ Radicals which are similar can be added and subtracted by using the distributive law. - 224 - Note that in this case the Similar radicals have the same bases are alike so we could have index number and the same number multiplied these by using the under the radical sign. multiplication law am°an= am+n. Are 3\/3 and —2\/3 similar radicals? (b (27)l/6= (33)1/6= 31/2= Wh 7 (C; (l6)1/6= (24)1/6= 22/3zléfiéz y (d) 61/3'121/2z 62/6.123/6= (62°123)l/6. Instead of multi- plying 62 by 123, factor these numbers to see if you have a perfect sixth power and so can reduce the radical. (3'2‘3‘2'2°2‘3°2°2°3°2°2'3)1/6= (35.26.22)1/6= 2(35.22)1/6= W 65. Yes because they have the same 66. Since 3\fig and -2\/5 are similar 66. index number of 2 and the same radicals they can be added by’ number under the radical sign usin the distributive law. of 6. 3 + -2\/Z =V'6(3+-2) = -1 or -' . Combine. ms +6 3-W3 Es? QC” 3V3}- 5ifv—V: v— c 3 ab + 5 2 - ab - 5 2a :1L2_ 2-15 V12+3VE= 66 .. 17 66 - gin/6 That is, (1- V3)(1+\/'3') = - 3+V3- 9 = 1-3 = -2. (e) 15 V6110 0+3V66.2V§3 = V— V- 110 - 14 10 In this product, there is no irrational number. (r) V6.16V6 = 2 - 48 = .46 In other words, we can rationalize (g) lZa-ll.Vab+2b a binomial which contains square roots by multiplying by a binomial (h) x+2‘ny+2x\fl§+2y\(x which consists of the same two terms but with one opposite sign. (1) x3 - y3 To rationalize \fii V3 -\/3 we must multiply both numerator and denominator by the same number and this number must be such that the new denominator will not be irrational or will not contain a radical. What would you multiply both numerator and denominator by here to rationalize the denominator? 76. V5 4’ v3 77. Rationalize y-Z- V5 -\/5 77.\/§(\/‘3’ +V5') =\{6 + V10 78. Rationalize 4 +55 '2 1 N3 78. Multiply both numerator and 79. Rationalize } -\ZZ denominator by l+\/§ so that the 2 V5 _ 2 new denominator will contain no radicals. (Mi/31mm z ka/E z (1- V3)(1+\/'3') 1 -\/6 2 all? -2 79. multiply both numerator and 80. Fractions should always be left denominator by 2\V§ + 2 . in their lowest terms - that is, the numerator and denominator -228- (3— WLLZ \[2—4-2) = 6+1+\/'2-2W should have no common factors. -_ " 4 _ Look at the answer to the last (2 V2 2)(2V2+2) hv— “ problem. Her both numerator and = 2 + 1+ V2 denominator have a common factor of . Reduce this fraction to its lowest terms. 80. common factor of 2 or -2. 8.1. Is __ 1+2 2 also a correct 2 2+4 Y2- : 1+2 y2 answer? Why? -2 81. Yes, because two factors have 82. Rationalize the following fractions. have been multiplied by -l. (a) 33/2 - 2V5 V6 (b) 1V3 - 2V1 V2 - N5 (CM/3 - 3V5 2\fl§-+11\E§ (d) 2 + 3V2 2 - 5V2 (6) x +33fE§‘+y v; ..v; 82. (a) :2 V212 Vi ° V; = 6M3-12y5 83. Earlier we stated that theVI \/6 V6 didn't exist because there is no __ _ real number such that when that =V3 - 2V5 that number is squared the result is .14., In fact, we went on to (b) g3! E-ZVEZSV ZflgL say that the nth root of a negative 2W3 ”M number is £123 a real number, when M )(v— 3) n is even. 10- 6+12 1 -8 _ __ " - What about the odd noot of a neg- VI; 16 V9 ative number? Is this a real 3 V10 .. 2 V6 + 12 V15 - 24 number? - O (c) (Yd-fiMZXE—WE) = For instance, can you find a real number which is the cube root of (2WV6M2V6-W2 > _8? 2! 35-10y124-12y4 26-20y E 2-3g was .. 16W: -8 -2 (d) g§+3V2232+§V22 __ (2-5V?)(2+5’V§) 10+31V2+1fl§ = 1+0 + 315/? u .. 25W 4‘5 - 229 - (e) W ._. (vs-mum) x x+’x x+x X - Y 83. Yes to both questions. —N is a real number when n is odd. It is n23 a real number when n is even. 84. (a) Zi\/§ (b) i (c) lZi (d) 5+6??- (e) 21% 85. real part is 5 imaginary part is 61 84. 85. 86. If we try to reduce v-4 by the same procedures we used a few frames back in reducing radicals, we obtain, VJJI =\/Z-l5ws = ‘VII'VE'= 2 -1. Let us define V-l = :1. Then 2 v: = 21. This is an ex- ample of an imaginary number. Imaginary numbers are the even roots of negative numbers. Express each of the following in terms of i. Reduce all radicals. (a)\/:§ (b) V3 (0) 3 V-13 (d)v2'5’ +\/-"3'6 (e)\f_-12 Note the answer to part d of the last frame. 5+6i consists of a real number plus an imaginary number. The real number is . The imaginary number is . A number such as 5+6i is called a complex number. A complexfinumber is a number of the form a+bi where a and b are real numbers. To add or subtract complex numbers, procedd as in any other addition or subtraction problem by combining similar terms. For example, the sum of 3+4i and 5+3i is found by adding the real parts and adding the imaginary parts. (3+“i)+(5+31) = 3+5+4i+31 = 8+i(4+3) = 8+7i. - 23o - Find the sum of (a) 3-61 and -2+Si (b) 2-i and -3-9i Find the difference of (o) 14461 and 7-21 (d) 51-6 and 2+8i 86. (a) 3-6i+(-2+5i) = l-i 87. In multiplication of complex (b) 2-i+(-3-9i) = -l-lOi numbers, we use the distributive (c) 4+3i-(7-2i) = -3+Si law as we would if we were dealing (d) 51-6-(2+8i) = -8—31 with real numbers. The commutative, associative and distributive laws hold for addition and multiplication of complex numbers. (3wi) (2.71) = 8?. 6-13i-28i2 88. Let us consider i2 to see if this can be expressed in another form. i =-V-l , therefore i2 = (\/:—)Z _[_1)1/232__ So -2812 = -28(-1) = Then 6.131.2812 = . 88. 6-l3i+28 = Bb-lBi 89. The product of two complex numbers can always be expressed as a complex number. Express each of the following as a complex number. (a) (2+9i)(-5-21) 2 (b) (3-51) (a) (31 V3 - 2)<2i\/‘5' + 3) (d) (V2 - iV'é’XBVE + 21%) 89. (a) -lO-49i-18i2 = 8—49i 90. Fractions which have complex numbers in the denominator must be ration. (b) (3—51)(3-5i) = 9—301+25i2= alized in the same way as fractions _16-301 which have irrational numbers in 2 the denominator. (c) 61 V23+5ill3.6 = -30+51 -6 = -36+5i 5 For example, 1+ 1 can be ration- 3- i (d) 3 V5.1 V12.2i2 V66 = alized by multiplying both num. erator and denominator by a number 18 - 2i\/6 which will make the denominator a -231- 90. multiply by the conjugate which 91. Rationalize in this case is 3+Qi. What can 3—fli be multiplied by so that the result will be a real number - that is, so that the product will not contain 1? 63?: 91. 1+ i = 2:131+12i2 = 92. -2:lji can be written as‘_'2_ +.lii - 1 i or as._.42 1 . 25 25 ‘* 2%“ ' You will note that this result is also a complex number or is of the form a+bi where a and b are real numbers. The real part of this number is ___, The imaginary part is . 92. real part is ~9/25 93. The quotient of two complex imaginary part is 13i/25. 93. §§EEigéz+uig _ 10+26i+1212 2 i 2 i - u _ 1612 -2+26:L____2_ _2_6. =__l 13. 20 ‘ 20 I 20 10 + 101 94. i” = “.12 ( -l)(-1)= 95. i5 - i”°i = (l) i = i 96 i — 4-12 - <1)<-1) 94. 95. 96. 97. numbers can always be expressed as a complex number. Express Efigi 2 i Let us consider the values of 13 as a complex number. and 1h. ‘we already know the values of i and 12, so we shall express i3 and i“ in terms of these quantities. i3 = iz'i = (-1) i = -i. L; i = = = . What is the value of 15? What is the value of i6? If you examine the powers of 1 further, you would find that all the odd powers of i are imaginary numbers. What kind of numbers will all the even powers of i be? -23 97. real numbers 98. i3 = -i, 15: i, 16: -1. i3+ui5-2i5= -i+u(i)-2(-1) = 2+3i 99. Of course it can. by any number is 0 regardless of whether the number is a real number or an imaginary number. 100. No. Any number which contains i or the even root of a negative number is not a real number. 101. real and imaginary 102. rational and irrational 103. the integers and the rational numbers which are ggt integers The rational numbers which are not integers are sometimes referred to as fractions or nonintegers. 0 L,- 98. Not only will even powers of i be real numbers, but even powers of i will either equal +1 or -1. Express as a complex number: i3+ui5.2i6 99. All real numbers are complex numbers. Consider the real number ~5. It can be expressed as -5+O°i. Can any real number be expressed as a+O'i? O multiplied 100. Are all complex numbers real numbers? Give an example to illustrate your answer. 101. Let us review the various kinds of numbers with which we have dealt. The largest set of numbers is the set of complex numbers. Complex numbers are composed of two kinds of numbers - the numbers and the numbers. 102. The real numbers are composed of two kinds of numbers - the numbers and the numbers. 103. The set of rational numbers is composed of two parts — the and the . 104. A chart of the numbers we work with would look like this. integers rational - nonintegers real - numbers irrational complex - numvers imaginary numbers Complete the following. (a) Complex numbers are numbers 104. 105. 106. (a) of the form a+bi where a and b are real numbers or are composed of a real part and an imaginary part. (b) even roots of negative numbers (c) Rational numbers are real numbers which can be expressed as the quotient of two integers where the denominator isnot 0. (d) real numbers which can't be expressed as the quotient of two integers. Note: in order to decide on the smallest possible classi- fication, you may have to change the form of the given number. (a) rational (b) real (0) complex A fraction with O in the denominator doesn't represent a number. (b) Imaginary numbers are (c) Rational numbers are (d) Irrational numbers are 105. State whether the numbers in the following are complex numbers, real numbers or rational numbers. Use the smallest possible classification. (a) ~2,Viza {1‘89 09 23/4 (b) V517. \2/-32. V371 (c)\/:E, O,\2[:§I 106. What symbol have we come across which doesn't represent any number - that is, what symbol isn't defined? Chapter 8 - Quadratic Equations ‘We will learn how to solve quadratic equations and problems which lead to quadratic equations in this chapter. 1. A quadratic equation in one variable has only one variable and contains the second power of this variable but no higher power of the variable. For example, 3x2: B—Qx is a quadratic equation in x. Is 4x2- 5x = O a quadratic equation in one variable? Why? 1. Yes because it involves the 2. Which of the following are quadratic second power of the variable but equations in x? no higher power. (a) u - 3x2: 7x (b) 5 + 2x2 (C) 4x(x2 - 3) = 7 (d) 9x2 = -9 (9) 4x2 = O 2. Parts a, d and e are quadratic 3. One of the theorems in algebra equations in x. states that an equation where the Part b is not an equation as an highest power of the variable is equation needs two equal quanti- n and n is a positive integer ties. will have precisely n roots. Part c will be 4x3-12x = 7 when the parentheses are removed and In a quadratic equation, the the highest power here is 3, so highest power of the variable is this isn't a quadratic equation. and so a quadratic equation You must remove parenthese and has precisely roots. fractions and collect terms before deciding what degree equation you have. 3. two a. Every quadratic equation is of the two form axz + bx + c = o. If the equation isn't given in this form, it can be put into this form. Put the equation 4 - 3x2 = 7x in the above form. - 234 - #. Either -3x2-7xfi4 = O or 3x2+7x-h 503:39b 0. 7 and c _235- 5. In ex2+bx+c = o, a stands for the coefficient of the squared term, b stands for the coefficient of the first powered term and c stands for the constant term or for the term which doesn't contain the variable. In the equation 3x2+7xa4 = 0, what are the values of a, b and c? -4. 6. In the equation 6x2-5 = 7x, what are the values of a, b and c? 6. First, set up the equation so 7. In the equation 4x2 = 9, what are that all the terms are on one the values of a, b and c? side of the equation equal to 0. In 6x2—5-7x = O, a and c = -5. In 7x-6x2+5 = O, a and c = 7. In 4x2-9 c = -9. In 9-4x2 c=9. - = 9 8...? E 5. O, a = O o m l 4. 6, b = -7, -6, b = 7 b = O and 8. We will first consider quadratic equations of the form ax2+c = O or where b = 0. - -4, b = O, and In equations of the form ax2+c = o, arrange the equation so that lx2 is equal to some number and then take the square root of both sides of the equation. For example, in hxz = 9, first find out what x2 equals. x2 = 9. If you wrote 2%, change it back to the fractional form. 2% is correct, but it is much more convenient to deal with fractions in taking roots. We have x2 =.§.. Now take the square root of both sides of the equation. There are two square roots of every complex number. In this case, x = +3/2 and also x = -3/2. When you square either of these numbers you get 9/h. Find the roots of 3x2 = 75. - 236 - 9. x2 = 25, x = 5, x = -5 10. 10. Do Egg substitutions in the 11. original equation - one when x = 5 and the other when x = -5 and see if you obtain the same number on both sides of the equation. ll. (a) x = 6 x = -6 12. (b)y=\V/39Y--3 (c) x= 5/3. x=-V'5'/3 (e) x2=-8, x=:V:§, 75:12:45 (d) x2 = - 16, x =‘:_ -l , x =‘1_4i (f)y=1vg ory=1 74 (g)y=+ -30ry=ii\/§ 12. 4x2 - 5x = O or 5x - hxz O 13. 13. x(4X—5) = O 14. 14. One of the numbers or else both 15. of the numbers equals 0. That is, if ab = 0, then a or b or both equal zero. The roots of this equation are often written x =‘1,5 meaning the roots are 5 or -5. How would you check to wee if 5 and —5 are the correct roots to the equation 3x2 = 75? Solve and check each of the following. (a; hxz - 1&4 - O (b = 27 9x2=5 (d) 2x2 + 32 = o (e) 4x2 + 32 = o (f) 7y2 = 2 (g) 8y2 + 24 = 0 We can only use this procedure when b = 0. That is, we can only use the above procedure when we have a term containing the var- iable to the second power and a constant term. Suppose we have a quadratic equation containing all three terms or one which contains a constant term of 0. First arrange the equation so that all the terms are equal to O. For example, in the equation 4x2 = 5x, we would write . Then if the quadratic expression is factorable, factor it. Thus 4x2 - 5x = 0 becomes . Consider this equation. x(4x-5) = 0 On the left side of the equation, we have the product of two numbers. On the right side of the equation, we have 0. What do you know about two numbers if their product is 0? Then in x(4x.5) = 0, either x = O 4x~5 = O. This means that either x = 0 or x = . - 237 - 15. x = S/U 16. To solve the equation x(x-3) = u, we should first ________. 16. Set everything on one side of 17. Note: we cannot set each factor the equation equal to O. in the given equation equal to some number in this case. Which numbers would we use? There are many possibilities of two numbers whose pro- duct is t. We can only use the principle of setting each factor equal to some number when the product of the factors equals 0. Now we have the equation x2-3x-4 = 0. Finish solving. 1?. (x-4)(x+l) = O 18. How would you check to see if these x-4 = O x+l = O are the correct roots to this x = 4 x = —l equation? 18. Substitute in the original l9. For x = 4, the left side of the equation. equation becomes “(h-3) which You must do two substitutions, equals 4(1) or 4. The right side one using x 3‘5 and the second of the equation is u and since using x = -l. the two sides of the equation are the same, x = h is a correct root of the equation x(x—3) = 4. For x = -l, the left side of the original equation becomes -l(-l—3) which equals -l(-4) or 4 and since this is the same as the right side of the equation, x = -l is a root of the equation x(x-3) = h. Solve and check. (a) 8x2 - 64x = O (b) 4x2 = 10x (c) 2x2 + 5x = 3 (d) 6(x2-l) + 5x = 0 (6) 5x(3x+4) = x+10 (f) 6x2 = llx+7 (g) x2 + 2ax - 3a2 = O (Solve for x in terms of a.) 19- (a) x = O, x = 8 20. There are many quadratic expressions (5) X = 0, x = 5/2 which do not factor easily, some (0) X = %, x = -3 which factor only when we use (d) Be sure to remove parentheses irrational numbers and some which and set equal to 0 before factor only when we use complex trying to factor. numbers. It would not be very 20. 21. 22. 230 24. - 238 - x = -3/2, X = 2/3 (6) AA 0?: H) VV x +2ax+a 2a Be sure to multiply and collect similar terms and set equal to 0 before factoring. X=2/59 3‘: -5/3 x=-%9 Xz7/3 (x+3a)(X-a) = 0 x+3a = 0 x-a x = -3a x = O SDI! \1.) ll ... I:- 4 and x - 3 = -4 Kit) ll 21. 22. 23. 2#. 25. convenient to solve all quadratic equations by factoring. We can use the first form we talked about to solve quadratic equations, but this form only works when the term containing x has a coefficient of 0. If we had (x.+a)2 = N, we could then take the square root of both sides of the equation and solve the resulting equations for x. For example, (x-3)2 = 16 is an equation of this sort. Taking the square roots of both sides of this equation, we get . x-3 = 1% is equivalent to having the two equations x-3 = h and x-3 = -u. Solving these two equations for x, we get . Our problem then is to put any quadratic equation in the form of (x_+a)2 = N. We can do this by the process of completing the square. Let us examine (x+a)2. This equals . x2+2ax+a2 is sometimes referred to as a perfect square trinomial, because it is the result of squaring a binomial. The coefficient of x in this expression is . Take half of this coefficient and square the result. What do you get? Note that this result of a2 is the same as the constant term in the expression x2+2ax+a2. a2 is the term which is necessary to complete the square of x2+2ax. The necessary constant term can always be found by taking half of the coefficient -239.- of the first powered term and squaring the result providing the coefficient of the squared term is 1. Complete the square of x2-6x. 25. The coefficient of x is -6. 26. Find the factors of x2-6x+9. The coefficient of x2 is 1. Therefore, take half the coeffi- cient of x which is -3 and square it getting 9. +9 completes the square of x2-6x. 26. (x-3)(x-3) 27. Is x2-6x+9 a perfect square trinomial? What must be true in order to have a perfect square trinomial? 27. Yes, because a perfect square 28. Complete the square of x2-3x and trinomial must factor into a then factor the resulting binomial multiplied by itself. expression. 28. 9/4 completes the square. 29. Then to solve the equation x2+2x-2 = o by the method of XZ-3X+‘g completing the square, we will first arrange the equation so that (X -‘%)(x -.g) the terms containing the variable are on one side of the equation and all other terms are on the other side of the equation. x2+2x-2 = 0 becomes . 2 29. x +2x = 2 30. In ogder to complete the square of x +2x we need ______, 30- *1 31. If we add a number to one side of an equation, we must add the same number to the other side in order to have equivalent equations. 2 x2+2x = 2 becomes x +2x+l = 2+1 or x2+2x+l = 3. Factoring the left side of the equation, we get (x+l)(x+l) = 3. This can be expressed as (xv-1)2 = 3. This is the form mentioned in frame 20 and so we can take the square root of both sides -240- of this equation. We get . 31. x + l = 1\[3 32. Solving these two equations, we will get the roots of the or equation x2+2x-2 = O. x + l =\/3 and x+l = -\fl3 The two equations are and . Therefore, the roots of the equation are and . 32. x+1 = 3 and x+l = .\/3 33. Check these by substituti in the original equation of +2x-2=0. roots are\/3 - l and\/§ + l 33. (V3.1)2+2( V3—l)-2 = 31+. To review, in solving an equation by the method of completing the 3-2 Wl-bZV-BZZ-Z = O and since square, we first arranged the this is the same as the right equation so that all the terms side of the origina equation, containing the variable were on V3.1 is a root of +2x—2 = 0. one side of the equation and all other terms were on the other Substitute in the original side. equation again using x = -‘(3—l. Next, we completed the square for the terms containing the variable being sure to add the same number to both sides of the equation. We then took the square root of both sides of the equation ob— taining 332 linear equation. Now solve these two linear equations for the values of the variable. Check both of these values to see if they are roots to the given equation. Solve byzcompleting the square: x -3x.5 = O. 34. X2-3x = 5 35. Remember we can complete the x -3x+9/4 = 5+9/4 (adding the square by taking half the coeffi- same number to both cient of the first powered term 2'2 sides of the equation) and squaring it only When the (X- 2) =.§g coefficient of the squared term a is l. x - 45- : 1E (taking the square 2 root of both sides) Sggpose we had the equation 2 -5x-6 = O and we wished to x - 2 = [23. or x - 2 = -\£2§ solve it by completing the square. NEE? 2 §i§§ 2 What could we do so that the X = 2 +‘% , x = - 2 +.% coefficient of x2 becomes 1 and we have an equivalent equation Be sure and check both of these. to 2x2-5x-6 = 0? 35. 36.£ 37. 38. 39. 40, - 241 - Divide both sides of the 36. equation by 2 getting £-gx-3=m {afix +._%=3 +§§ 37. aft-=3- 2V2? ”5V?- emu- g,x x=-1: u u Check both of these in the original equation x ~3x-5 = 0. Make the coefficient of x2 equal to l by dividing both sides of the equation by a. 38. Arrange the equation so that the terms containing x are on one side of the equation and all other terms are on the other 39. side of the equation. 2 b _ c X+EX"--a- x2 + E'x + b = _.2 +.;b2 a he a 4&2 Express the left side as the #0. square of a binomial and combine terms on the right side of the equation. 2 “3 uaz Taking the square root of 41. both sides of the equation, we 2 getx+_§=+ -4 zafi _ b b -4ac x — - + 23' " 2a Solve x2 -13x -3 = O by completing the square. If we solve thg general Quadratic equation of ax +bx+c = O by completing the square, we will derive a formula that can be used to solve every quadratic equation. To solve ax2+bx+c = 0 by completing the square, we would first have to . Do this. Before we complete the square, we would have to Do this and then complete the square. What has to be done next? Do this. Complete the solution. Inx=--—b- _+_ b-L‘ac,the 2a 2a denominator of both fractions on the right hand side of the equation is the same, so we may ...b + Vbz-ltac 2a This represents the two roots to a quadratic equation where a is the coefficient of x2, b is the coefficient of x and c is the write x = 41. 42. 43, First arrange the terms of the #2. equation so that all terms are on one side of the equation equal to 0. Then choose the values of a, b and c for this particular equation. -5, x2-5-3x = 0, so a and c l, b -3 Quadratic formula: ..h-L VbZ-ltac 2a So in this case, -(-3) 1W 43. X x: + 2(1) x = 3' 29 , which can be written ”25% Xaiaiéé 2 2 3x2-5—Qx = 0 so a = 3, b = .h 44. and c = -5. x ... ..(Jt) iVTFl - (TF3 :5)" 2(3) ...—H—‘glCEXLgBCZ—fi constant term after all the terms are put on one side of the equation equal to O. \f 2 One root is x = 'b+ b -hac 2a and the other root is = -bJYb2-4ac 2a To solve x2 = 5+3x using the quadratic formula, we would first . Do this. It doesn't make any difference in what order the terms are arranged in choosing the values of a, b and c as long as‘gll the terms are on one side of the equation equal to O. This is because a is always the coefficient of the squared term, b is always the coefficient of the term to the first power, and c is always the constant term or the term which doesn't contain the variable. Using a = l, b = -3 and c = -5, solve for the roots of the equation 5+3x using the quadratic formula. In this case\/§9 cannot be reduced. If we obtain a radical which can be reduced, we do so in order to get a fraction which is in its lowest terms. Solve and check using the quadratic fomflae 3x2-5 = #x If you haven't reduced the radical and then reduced the fraction to its lowest terms, do so now. - 243 - 44. x =.&_I_§k[:§ , x = h-2\/l9 45. Using the quadratic formula, 6 3 solve 3y(y-l) = y-5. Both of these fractions have a common factor of 2 in both numerator and denominator. Reducing fractions, we get 21W? 3 Don't forget to do the check. x = for the two roots. 45. 3y2-4y+5 = 0 so a = 3, b = -u 46. Solve the following equations. and c = 5. Use factoring where possible. Check = 4-4) :VIEJLIBKST (a) 1 a 0 2(3) y_L+.t\/'.'.'EE y_l+: 2i\/.'L‘I (b)3-2X2=21X .._________. , _ 6 6 (c) 3x(x#2) = 2x-l Y =.EJEIE>€EE (d) 2x(x~6) = (X-3)(X-2) 3 (a) 453-100 = o (f) 9y2+6y+4 = o + 46. (a) x = :-;§:£1;3: 47. Be sure you do the check parti- cularly in parts a, c and f as or x = (..l g; iV'3)/2 this will give you further work on handling irrational and (b) Ybu can factor this. complex numbers. 3x(4xp7) = 0, so x = O, x = 7/4. Some equations will involve fractions. (0) Remove parentheses and com— It is usually easier to obtain bine terms getting 3x2+4x+l=0. an equivalent equation which You can factor this getting contains no fractions. x = -l, x = -1/3. Remember - If you multiply both (d) Remove parentheses and sides of the equation iambine terms, getting by (x-Z), you obtain ~7x~6 = 0. equivalent fractions Solve using the formula. only if you haven't 7 + multiplied by O. In x = " this case, then, x f 2. 2 (e) You can solve this by If you have an equation with factoring or you can set several letters in it, choose the the equation up as 4x? = 100 variable you wish to solve for and and then take the square then determine the coefficients root of both sides of the of that variable. equation. Roots are 5 and -5. For example, if you are to solve for x in the equation a x2—3a2x+5=0, you need to find the values of .y '11.! 1i! Ill. - 244 - (f) solve using the formula “-6in 18 ° radical and put the fraction in lowest terms. y=uiiy3 3 47. b = .3a2 and c = 5. Reduce the 48. Now substitute these in the formula, getting x ._. 3a2 :W-Malhw) 2(a“) You can combine and simplif" as you did in the other problems. 48. (a) LCD is 15(x-4)(x-2). Multiplying both sides of the equation by this gives 15(XPB)(X-2)-15(X-l)(X-4)=2(X-4)(X-Z). Roots are 7 and -l. (b) LCD is (2x-3)(x-5)(3x+2). Multiplying both sides of the LCD gives (5xs8)(3x+2)-(6xw4)(X~5)=(4x+4)(ZX-3). The resulting equation must be solved using the formula. +\,—— Roots are x = '16 g’ 192 . 49. The fraction can be reduced, so the roots are x = —8 + 4\/—. (c) Solve using the uadratic formula with a = 3 5, b = -10 and c =\[§, + 0‘- Roots are x = 10 "VAC . 6/3' a, b and c in order to use the quadratic formula. 4 In this case, a = a , b = and c = . Solve for x in each of the following and check. (a) x- xel_ .:E’ x-Z (b) .5xe8 _, 6xfi4 = 4x+4 2x2‘13XI15 6X2-5XP6 3x2-13x.10 (c) 3V3 x2 +\/'5‘ = M (d) xz+2xy+4y2 = (e) xZ + 16y“ = 4x312 (f) “DC-l __ X411 = 3 x+2 (g) 2-X _3x-2_ -6 x-3 x+3 x2—9 In part a of the last frame, we multiplied both sides of the equation by l5(x.4)(x—2). This means that 4 and 2 cannot be roots of the equation because our new equation is not equivalent to the given equation —if x = 4 or if x = 2. Why don't we obtain an equivalent when x = 4 or x = 2? What values of x can't be roots of the equation given in part b? of the equation given in part f? of the equation given in part g? - 245 - Reducing the radical and the fraction, we obtain X z 5.2. 10 3\/§ (d) Solve using the formula, with a = l, b = 2y and c -4y2. -2 1 Roots are x = . 2 Reducing the radical and the fraction gives x = -y+ yi f3. (6) Solve using the formula with a = l, b = —4y2 and c = 16y“ . 2 +\/ E Roots are x = 4y' " -48y . 2 Reducing the radical and fraction gives x = 2y_ 8iy2\/3. (f) LCD is (x+2)(x+4). Multiplying both sides of the equation by this gives (4x-l)(x+4)-(x+1)(x+2)=3(x+2)(xsu). When we multiply and collect terms, we do 222 have any squared terms, so this is a linear equation and has only one root. Root is x = -5. 'cgy LCD is (x-3)(x+3). Multiplying both sides of the equation by this gives (Z-X)(x+3) - (Bx—2)(X-3) = -6 Solving this, you obtain x =-:, x = 3. However, note that the 3 doesn't check in the original equation because it makes one of the denominators equal 0 and we can't divide by O. Therfore, there is only one root to this equation, namely x = -§. 49. In part b, 3/2, 5 and -2/3 50. The roots of the equation in part can't be roots. e of the last frame are given as In part f, -2 and -4 can't be + roots. x = 5'- \/lO , In partg ,3 and -3 can't be roots. 3\/F In order to rationalize this fraction, we would have to Do the rationalization. 50. 51. 52. 53. 5h.[Ex-4)l/é]2 = x-4 - 246 - Multiply both numerator and 51. denominator'QY\/§. szvfi' \5=5V'§:V§5 n5 V3 5 Reducing the radical and the fraction ives \5:v§ 3 a = 1, b = 4y, 0 = 16y2-16. 52. X = ..pr 1V16y2-4(1)(16y2-167 53 2(1) y = ..l'ry i v-48y2+61+ 2 You must have factors under the 54. radical in order to reduce it. xz-W:VRLrAM 2 x = -4y 1 4 VLF-63r2 2 -2y i 2 V4~3y2 >4 ll 55. Usually fractions involving radicals are rationalized as then they can be expressed as a rational number multiplied by an irrational number. What values of a, b and c would we use to solve x2+hxy+16y2 = 16 for x using the quadratic formula? Using these values, solve for x in x2+1+xy+l6y2 = 16 using the quadratic formula. If you haven't reduced the radical, do so now, and then reduce the fraction, if possible. Remember, to reduce a radical you must have what kind of quantities under the radical? At this point, we have solved for x in terms of y or in other words, the values of x depend on the values of y. Let us now consider irrational equations. An example of an irrational equation is x- + 3 = 0. The solution of irrational equa- tions depends on a law of exponents which involves raising a quantity to a power. We know that (al/n)n = a. Therefore, if we had (\/x~E)2, this would equal _______, What can we do to the equation \/x-E + 3 = 0 so that if we squared both sides of the e ua- tion, we would have only \[x-h as the quantity to be squared? 55. 56. 57. 58. 59. - 247 - Add —3 to both sides of the 56. equation getting with-3 X‘L" = 9 570 fig. _If x = 13, Vx+4 + 3 becomes \/l3:E + 3 which becomes 6. This is not equal to the number on the right side of the equation. 3 V2x-7 = 3 59. 9(ZX-7) = 9 X 2 What do you have to do now to see if this value of x is a root to the equation? Be sure you do this. Arrange the equation so that 60. the radical is on one side of the equation and all other terms are on the other side. -2‘V 2x+5 = l u(2x+5) = 1, x = -19/4. No solution as this doesn't check. If we now square both sides of the equation x- = -3, we obtain . From x-h = 9, we find that x 13. When both sides of an equation are raised to the samegpower, we do not always_get equivalent equations. Therefore, we must always check the value or values we obtain to see if they are roots of the original equation. Is x = 13 a root of\/x-H + 3 = 0? Therefore, \/x—u + 3 = 0 has no solution. To solve 3\¢2x- - 5 = -2 for x, we would first arrange the equa- tion so that the term containing the radical is on one side of the equation and all other terms are on the other side. Then when we raise both sides of the equation to the same power, we will no longer have an irrational equation. Arrange the equation in this way, square both sides and finish solving. Remember when you raise both sides to the same power, you do get always get equivalent equations, so you must always check the values of the variable to see if they are roots of the equation. x = 4 is a root to the equation 3V2x—7 - 5 = _2. 3 - 2\/2x+5 = 4. If you had the equation Solve for x: ‘2/2xfi3 = 2, what could you do in order to get an equation which doesn't contain a radical? 60. 61. 62. Vx-Z = _.2 +\/3x-2 63. - 248 - .Raise both sides of the equation to the third power or cube both sides of the equation. cubing both sides of the equation, we get 2x+3 = 8. Solving, we get x = 5/2 and this .ig a root as it checks in the original equation. Squaring both sides of the equation x—Z = (-2 +\/3x--2)2 x-Z = {—2 +\/3x—2)(-2 +\/3x—2) X-Z = 4 — 4‘d3x-2 + 3X - 2 By repeating the same process 64. we used before. Arrange the equation so that the term containing the radical is on one side of the equation and all other terms are on the other side. Then square both sides. -2x-J+ = J+ V3x-2 Squaring both sides of this equation, we get 4x2+16x+16 = 16(3x—2). This gives us the quadratic equation 4X2 - 32x + #8 = 0. Solving this, we obtain x = 6, X320 Only x = 2 checks, so x = 2 is the only root of the given equation. 61. 62. 63o 65. Solve \3/2x+3 = 2 for x. The equation \/x-2 -‘¢3x+2 = -2 has two radicals in it. However, we can use the same principle to get an equation which doesn't contain any radicals. First arrange the equation so that gag radical is alone on one side of the equation and all other terms are on the other side. Then since we have square roots, square both sides of the equation. Do this being sure to square the complete side of the equation. Note that the right side of the equation before it was squared was a binomial. Therefore, when it is squared, you are multiplying two binomials together. Now we have an irrational equa- tion which only contains one radical and we are interested in getting an equation which doesn't contain any radicals. How can we do this? Complete the solution. After squaring both sides of the equation ~2x—h = -4\/3x-2, we obtained 4x2+16x+16 = 16(3x-2) . We can see that when terms are combined, we will still have a squared term. This means that we have a quadratic equation to solve. What methods do we have of solving quadratic equations? - 249 - 65. factoring, completing the 66. square, and the quadratic formula. 66. All the terms on one side of 67. the equation equal to O. 67. x2-8x+12 = o 68. In the solution of quadratic equations by the methods of factoring and the quadratic formula, how must the equation be arranged? Thus in this case, we get 4x2-32x+48 = o. In this equation, the left side of the equation has a common factor of 4. We obtain an equivalent equation when we divide both sides of an equation by a nonzero number, so we can divide both sides of this equation by 4, getting . We don't have to do this, but you can see that it will be easier to factor this equation or to solve it using the quadratic for— mula than to solve the equation given in frame 67 by either method as the coefficients are so much smaller. In fact, we could have divided both sides of the equation by a common factor when we had the equation _2x-’4 = J4 V3x-2. Don't forget to check the values 2% x you obtain in solving -8x+12 = O in the original equation which you were given in frame 62 to solve. What procedure would you do first in solving \f1512x,+ 7 —3\/x+3 = 0? 68. Arrange the equation so that 69. one radical is by itself on one side of the equation and all other terms are on the other side. Then square both sides of the equation. 69. Val—72x = 3 Vx-a-E .. 7 7o. Squaring both sides of the equation we get lh+2x = (3 xi - 7)( x+ - 7) 1144-2): = 9(X'9’6) - 2+2 + “'9 Do this procedure. Now what would you do? -250... 70. Repeat the same process. 71. -7x-89 = -42\J§¥8 Squaring both sides of this equation, we get 49x2+12u6x+7921 = 176u(x+6). This becomes 49x2-518x-27o3 = o 72. Use the quadratic formula or 71. Do this and obtain an equation which doesn't have any radical. 72. What method would you use to 73. completing the square because it would be too difficult to find suitable factors if there are any? Check these values in the original equation to see if they were roots to the given equation. 73- 74. (a) Vx-l = x - 3 x-l = x2-6x+9 -7X+10 = O The only root is x = 5. (b) VBX-S = 2 + VX- 3X-5 = 4 + 2 Vic-3 + x-3 2x-6 - 2 \/x-3 Dividing both sides of the equation by 2 we get x-3 = x-3. x2-6xv9 = x-3 x2-7x+12 = 0 Solution is x = 30 74. 75. solve this quadratic equation? Why? After you obtained values of x for the equation 49x2-518x-2703 = o by using the quadratic formula, what would you have to do? You must check both roots as you may have a case where there aren't any roots as well as the cases where there is one root or where there are two roots. We will not solve this quadratic equation as we are primarily interested in the procedure by which irrational equations are solved. If you need the practice in using the quadratic formula, you can solve this equation. Solve the following. (a)\/x-l - x + 3 = O (b) VBX-S - VX-B = 2 (c)\/2x2-5x-7 - 2 Vx+l = 0 (d) Vx—l - V7x+l + V2x+g = 0 Set us the equations needed to solve the following problems. Complete the solutions. (a) The product of two numbers is 24. Twice the sum of the two numbers less three times one of the numbers equals two more than the other number. Find the two numbers. (b) Two numbers differ by'#. The square of one of the numbers is 10 less than the other number. Find the two numbers. - 251 - (C) /2X2-5X-7 = 2 /;:I 2x2-5x-7 = u(x+1) 2x2-9x-ll = 0 Roots are -l and ll/Z as both check in the original 6 uation. (d) x-l = / x+l - /2x+g x-l = ( 7x+l - é2x+§§§ [fx+1 - /§E$8) x-l = 7x+l - 2 7x+l 2x+ + 2x+6 -8x—8 = -2 /(7x+1)(2x+6) 6nx2+128x+64 = 4(1ux2+44x+6) Roots are 5 and l as both check in the original equation. 75. (a) Let x = one number let y = second number xy = 24 2(X+y) - x = y+2 numbers are 6 and 4 or are -6 and -4, (b) Let n = one number Then n-4 = other number n2 = n—4 + 10 numbers are 3 and -l or are -2 and -6. (0) Let w = width Then 2w-3 = length w(2w-3) = 10b dimensions are 8 and 13. Note that we do not use the negative value here as a negative dimension has no meaning. (d) Let w = width Then w+3 = length wore) -4 = (numb dimensions are 9 and 12. Again we do not use the negative Value as a negative dimension has no meaning. (C) (d) The area of a rectangle is 10h sq. in. The length is 3" less than twice the width. Find the dimensions. The length of a certain rectangle is 3 more than the width. If the length is increased by l and the width decreased by l, the area is decreased by 4 sq. units. Find the dimensions. 50 9. 10. ll. 12. 13. 14. 15. 16. 17. 18, 19 Test A 34.8 {-.0004 = 2 +222:— Division is the inverse of . Given the numerals -6, if, 4, 3.3, V5, vszfiii, 314,77. (a) the integers are . (b) the fractions are . (c) the rational numbers are . (d) the irrational numbers are . (e) the real numbers are . If 3 ¢ 0 and b ¢ 0, then 0 = -3a2b fifl- O - The gum of the absolute values of -5, 2, -3 and The absolute value of the sum of -2, -3, 15 and Add: hxe3y+8-lOc and -x-5y+hc. Subtract —3p+5mp2q+4r from 2p_4m-6q+4r. (2a - 5b)2 = (6x+5y)(BX-4y) = - 2 7b‘IL _ a :%%fi+9m2= 2 9xy Divide 36x-2x2+8x4-10 by 2x2—3x+5. Show-work. [2p — 59.-.;ng + 5<3r-q>) = ...—...... In 4x3+5c (a) 3 is called an (b) 4 is called an (e) 4x3 is called a b-4le lfl 4H =__l_ - 252 - -6 is -12 is -253- 20. If a, b and c represent real numbers, state the law or prOperty of real numbers illustrated in each of the following. (a) b(a+c) = (a+c)b (b) c + (b+a) = (c+b) + a (c) a(b+c) = ab + ac (d) b°O = O (e) c'l ll 0 (f) O + a = a Test B 1. Factor the following. (a) (b) (C) (d) (e) (f) (g) (h) (i) 32ab - 8b2 ab + 6c + 3a + 2bc 16x“ - 81y” 25a2 - 49b2 ex-hfl- 16x2 - 40x + 25 4x2 + 29xy - zuyz 54y” - 250y x2 - 13x - 12 2. Perform the indicated Operations and simplify. Show work. (a)fl (b) (e) (d) (e) (f) +E%._ 8y28 + xy6x2 3x+ __£L_.._.2;:;a a - 2 2 - a 2a-6 2 a2-9 a i 3 (2a - 5)2 - 2(2a + 1) (2a - 5) - 4 (a - 2)(a + 2) :_ (a2 — ulflb + 3) ab(a2 - 9) ab - 3a + 6b — 18 O 4x - 16 12 Z-x-O-Z 3. Solve the following equations for x. Show work. (a) (b) (a) If you can buy b books for c cents, represent the cost of one (b) George's age now is 2a + 1 years. as George's age was 3 years ago. Represent Mary's age now. (0) A car travels d miles in h hours. 1 u 1 E=§'&*w 4x - 1_ 2x + 1 Ex + 1 2x - 1 book. Represent the cost of r books. - 254 - Mary's age now is twice as much Represent how far it travels -255- in x hours. 5. (a) Write as a single fraction. (b) Is this any different from tie+dne2oie2 tdeJ uuh1ofle4 Test C Show work on all problems. I. Solve by substitution: 2x - y = 5 + By = 2. Solve by addition or subtraction: _2x + X = 3. a. 2x2 + 3y: Evaluate: (a) (-8)2/3 (b) {-8)'2/3 (c) 4‘1/2 + to + 42 Perform the indicated Operations and simplify. Express all results with positive exponents. (a) (~2x3y>(xy)2 8 . (b) 35- : x2y y (C) (~3x1/2y-3)(2Xl/3y1/2) (d) (81X-3y1/2)-1/2 (e) x"2 + 2X”1 (1 + 2):)"1 Perform the indicated operations and simplify. (a) 2V'3‘(3\/5 - 5 + 5V1?) (b) (2\/§ - u\/Z)(3\/3 - 2\/3) (C)Vl—5_5-v3+V-- -\/2— (dHZE-t ME 2V' ..v2 (9) x - y .§§I up only. In a certain rectangle the length is 2 feet less than 5 times the Width. The number of square feet in the area is 7 more than the number of feet in the perimeter. Find the dimensions. - 256 - Test D 1. Solve: x2 = x + 6. 2. Solve: 2x2 - 6x + 3 = O. 3. Solve: ‘V3x+7 - x = l a. Solve: 2X + x :1 = 9 5. Factor: (a) x3 + 6x2 - 7x (b) 16x“ - yu (c) ax + bx - 2a - 2b 6. (2a - 3b)2 = 7. Express in the form a+bi where a and b are real numbers. 2 - i 5 + 1 X2 3 . 8. Simplify: (.... - 4)(—1——) . (25 - 2) y2 X + 2y 0 y 9. Express as a single fraction: 3 — X + 2 3x 10. Express with positive exponents. Simplify. (a) be“? + y'zrl ll. (a) O 12. Simplify: (3(/§«+2+\/§)(2\/§ - 3\/§). 13. Each of r couples has 2 children and each of 5 other couples has 3 children. How many children do the r+s couples have? 14. The difference between the square of a positive integer and 6 times the integer is 16. Find the integer. -257...