Skip to content
Textbook Guides

Solution Manual for Finite Mathematics with Applications In the Management, Natural, and Social Sciences, 12th Edition

Solution Manual for Finite Mathematics with Applications In the Management, Natural, and Social Sciences, 12th Edition is the best resource for textbook solutions, helping you tackle even the hardest problems.

Preview 16 of 803 pages

See the first 16 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    S OLUTIONS M ANUAL S AL S CIANDRA Niagara County Community College M ATHEMATICS WITH A PPLICATIONS AND F INITE M ATHEMATICS WITH A PPLICATIONS IN THE M ANAGEMENT , N ATURAL , AND S OCIAL S CIENCES T WELFTH E DITION Margaret L. Lial American River College Thomas Hungerford Saint Louis University John Holcomb Cleveland State University Bernadette Mullins Birmingham-Southern College

  2. Page 2
  3. Page 3
    Read page 3 as text

    Contents Chapter 1 Algebra and Equations 1 Chapter 2 Graphs, Lines, and Inequalities 47 Chapter 3 Functions and Graphs 94 Chapter 4 Exponential and Logarithmic Functions 160 Chapter 5 Mathematics of Finance 198 Chapter 6 Systems of Linear Equations and Matrices 234 Chapter 7 Linear Programming 316 Chapter 8 Sets and Probability 422 Chapter 9 Counting, Probability Distributions, and Further Topics in Probability 453 Chapter 10 Introduction to Statistics 490 Chapter 11 Differential Calculus 521 Chapter 12 Applications of the Derivative 603 Chapter 13 Integral Calculus 677 Chapter 14 Multivariate Calculus 751

  4. Page 4
    Read page 4 as text

    1 Chapter 1: Algebra and Equations Section 1.1 The Real Numbers 1. True. This statement is true, since every integer can be written as the ratio of the integer and 1. For example, 5 5 1  . 2. False. For example, 5 is a real number, and 10 5 2  which is not an irrational number. 3. Answers vary with the calculator, but 2, 508, 429, 787 798, 458, 000 is the best. 4. 0 ( 7) ( 7) 0      This illustrates the commutative property of addition. 5. 6( 4) 6 6 4 t t     This illustrates the distributive property. 6. 3 + (–3) = (–3) + 3 This illustrates the commutative property of addition. 7. (–5) + 0 = –5 This illustrates the identity property of addition. 8. 1 4 ( 4) ( ) 1     This illustrates the multiplicative inverse property. 9. 8 + (12 + 6) = (8 + 12) + 6 This illustrates the associative property of addition. 10. 1 ( 20) 20     This illustrates the identity property of multiplication. 11. Answers vary. One possible answer: The sum of a number and its additive inverse is the additive identity. The product of a number and its multiplicative inverse is the multiplicative identity. 12. Answers vary. One possible answer: When using the commutative property, the order of the addends or multipliers are changed, while the grouping of the addends or multipliers is changed when using the associative property. For Exercises 13–16, let p = –2, q = 3 and r = –5. 13. ( ) [ ] [ ] ( ) 3 5 3 2 5(3) 3 2 15 3 13 39 p q − + = − − + = − − + = − = − 14. ( ) ( ) ( ) 2 2 3 5 2 8 16 q r − = + = = 15. q r q p   3 ( 5) 2 2 3 ( 2) 1 + − − = = = − + − 16. 3 3(3) 9 9 3 2 3( 2) 2( 5) 6 10 4 q p r = = = − − − − − + 17. Let r = 3.8. APR 12 12(3.8) 45.6% r    18. Let r = 0.8. APR 12 12(0.8) 9.6% r    19. Let APR = 11. 12 11 12 11 12 .9167% APR r r r r     20. Let APR = 13.2. 12 13.2 12 13.2 12 1.1% A PR r r r r     21. 3 4 5 5 3 20 5 17 5 12            22. 2 8 ( 4) ( 12)     Take powers first. 8 – 16 – (–12) Then add and subtract in order from left to right. 8 – 16 + 12 = –8 + 12 = 4 23. (4 5) 6 6 1 6 6 6 6 0           

  5. Page 5
    Read page 5 as text

    2 CHAPTER 1 ALGEBRA AND EQUATIONS 24. 2(3 7) 4(8) 4( 3) ( 3)( 2)       Work above and below fraction bar. Do multiplications and work inside parentheses. 2( 4) 32 8 32 24 4 12 6 12 6 6               25. 2 8 4 ( 12)    Take powers first. 8 – 16 – (–12) Then add and subtract in order from left to right. 8 – 16 + 12 = –8 + 12 = 4 26.   2 (3 5) 2 3 13          Take powers first. –(3 – 5) – [2 – (9 – 13)] Work inside brackets and parentheses. – (–2) – [2 – (–4)] = 2 – [2 + 4] = 2 – 6 = –4 27.   3 2 ( 2) 16 2( 3) 64 1       Work above and below fraction bar. Take roots. 3 2 ( 2) ( 4) 2( 3) 8 1       Do multiplications and divisions. 3 1 2 2 6 8 1     Add and subtract. 3 12 1 14 2 2 2 2 7 1 7 7 7          28. 2 2 6 3 25 6 13   Take powers and roots. 36 3(5) 36 15 21 3 7 36 13 49       29. 2040 189 4587 , , 27, , 6.735, 47 523 37 691 30. 187 385 , 2.9884, 85 , , 10, 63 117  31. 12 is less than 18.5. 12 < 18.5 32. –2 is greater than –20. –2 > –20 33. x is greater than or equal to 5.7. 5.7 x  34. y is less than or equal to –5. 5 y   35. z is at most 7.5. 7.5 z  36. w is negative. 0 w  37. 6 2    38. 3 4 .75  39. 3.14   40. 1 3 .33  41. a lies to the right of b or is equal to b . 42. b + c = a 43. c < a < b 44. a lies to the right of 0 45. (–8, –1) This represents all real numbers between –8 and –1, not including –8 and –1. Draw parentheses at –8 and –1 and a heavy line between them. The parentheses at –8 and –1 show that neither of these points belongs to the graph. 46. [–1, 10] This represents all real numbers between –1 and 10, including –1 and 10. Draw brackets at –1 and 10 and a heavy line between them. 47.   2, 3  All real numbers x such that –2 < x ≤ 3 Draw a heavy line from –2 to 3. Use a parenthesis at –2 since it is not part of the graph. Use a bracket at 3 since it is part of the graph.

  6. Page 6
    Read page 6 as text

    SECTION 1. 1 THE REAL NUMBERS 3 48. [–2, 2) This represents all real numbers between –2 and 2, including –2, but not including 2. Draw a bracket at –2, a parenthesis at 2, and a heavy line between them. 49.   2,   All real numbers x such that x > –2 Start at –2 and draw a heavy line to the right. Use a parenthesis at –2 since it is not part of the graph. 50. (– ∞ , –2] This represents all real numbers less than or equal to –2. Draw a bracket at –2 and a heavy line to the left. 51. 9 12 9 (12) 3        52. 8 4 8 (4) 4      53.   4 1 14 (4) 15 4 15 19               54. 6 12 4 (6) 16 6 (16) 22              55. 5 5 5 __ 5 5 5   56. 4 4 4 4 4 4      57. 10 3 3 10 7 7 7 7 7 7     58. 6 ( 4) 4 6 10 10 10 10 10 10       59. 2 8 2 8    6 6  6 6 6 6  60. 3 5 3( 5)    3 –5 15   3 5 15  15 15  61. 3 5 3 5   2 3 5   2 2  2 2   62. 5 1 5 1     4 5 1   4 6 4 6  63. When a < 7, a – 7 is negative. So 7 ( 7) 7 a a a       . 64. When b ≥ c , b – c is positive. So b c b c    . Answers will vary for exercises 65–67. Sample answers are given. 65. No, it is not always true that . a b a b    For example, let a = 1 and b = –1. Then, 1 ( 1) 0 0 a b       , but 1 ( 1) 1 1 2 a b        . 66. Yes, if a and b are any two real numbers, it is always true that . a b b a    In general, a – b = –( b – a ). When we take the absolute value of each side, we get ( ) a b b a b a       . 67. 2 2 b b    only when b = 0. Then each side of the equation is equal to 2. If b is any other value, subtracting it from 2 and adding it to 2 will produce two different values. 68. For females: | 63.5 | 8.4 x   ; for males: | 68.9 | 9.3 x  

  7. Page 7
    Read page 7 as text

    4 CHAPTER 1 ALGEBRA AND EQUATIONS 69. 1; 30062007 70. 8; 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015 71. 9; 2006, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015 72. 4; 2008, 2010, 2013, 2015 73. 4; 2006, 2007, 2009, 2011 74. 8; 2006, 2007, 2009, 2010, 2011, 2012, 2013, 2014, 2015 75. |10.6 14.9 | | 4.3 | 4.3 − = − = 76. | 63.1 ( 8.0) | | 71.1| 71.1 − − = = 77. | 1.0 63.1| | 64.1| 64.1 − − = − = 78. |10.6 ( 5.7) | |16.3 | 16.3 − − = = 79. | 5.7 ( 8.0) | | 2.3 | 2.3 − − − = = 80. | 1.0 ( 5.7) | | 4.7 | 4.7 − − − = = 81. 3; 30062010, 2015, 2016 82. 7; 2010, 2011, 2012, 2013, 2014, 2015, 2016 83. 6; 2010, 2011, 2012, 2013, 2014, 2015 84. 3; 2014, 2015, 2016 Section 1.2 Polynomials 1. 6 11.2 1, 973,822.685  2. 11 ( 6.54) 936,171,103.1    3. 6 18 289.0991339 7         4. 7 5 .0163339967 9        5. 2 3  is negative, whereas 2 ( 3)  is positive. Both 3 3  and 3 ( 3)  are negative. 6. To multiply 3 4 and 5 4 , add the exponents since the bases are the same. The product of 3 4 and 4 3 cannot be found in the same way since the bases are different. To evaluate the product, first do the powers, and then multiply the results. 7. 2 3 2 3 5 4 4 4 4     8.         4 6 4 6 10 4 4 4 4         9. 2 5 2 5 7 ( 6) ( 6) ( 6) ( 6)         10. 5 6 5 6 11 (2 ) (2 ) (2 ) (2 ) z z z z     11.       7 4 4 7 28 5 5 5 u u u        12.         4 3 5 3 20 23 6 6 6 6 (6 ) y y y y y         13. degree 4; coefficients: 6.2, –5, 4, –3, 3.7; constant term 3.7. 14. degree 7; coefficients: 6, 4, 0, 0, –1, 0, 1, 0; constant term 0. 15. Since the highest power of x is 3, the degree is 3. 16. Since the highest power of x is 5, the degree is 5. 17.     3 2 3 2 3 2 5 4 8 x x x x x x           3 3 2 2 3 4 2 ( 5 8 ) x x x x x x        3 2 13 x x x     18.     3 2 2 5 7 4 8 2 p p p p        3 2 2 4 ( 5 8 ) (7 2) p p p p         3 2 2 4 3 9 p p p     

  8. Page 8
    Read page 8 as text

    SECTION 1.2 POLYNOMIALS 5 19.     2 2 4 3 8 2 6 2 y y y y           2 2 4 3 8 2 6 2 y y y y         2 2 4 3 8 2 6 2 y y y y          2 2 4 2 ( 3 6 ) (8 2) y y y y         2 6 3 6 y y     20.     2 2 7 2 5 3 2 6 b b b b          2 2 7 2 5 3 2 6 b b b b              2 2 7 3 2 2 5 6 b b b b        2 4 1 b   21.     3 2 3 2 2 2 4 3 2 8 1 x x x x x           3 2 3 2 2 2 4 3 2 8 1 x x x x x         3 2 3 2 2 2 4 3 2 8 1 x x x x x            3 3 2 2 2 2 2 8 (4 ) ( 3 1) x x x x x         2 6 4 4 x x     22.     3 2 2 3 9 11 8 4 10 6 y y y y y            3 2 2 3 9 11 8 4 10 6 y y y y y          3 2 2 3 9 4 ( 11 10 ) (8 6) y y y y y         3 2 3 13 21 14 y y y     23.   2 9 2 6 1 m m m        2 ( 9 ) 2 9 (6 ) ( 9 )( 1) m m m m m        3 2 18 54 9 m m m     24.   2 2 4 6 8 a a a     2 2 4 2 ( 6 ) 2 (8) a a a a a     3 2 8 12 16 a a a    25.   2 (3 5) 4 2 1 z z z        2 2 3 2 2 3 2 (3 ) 4 2 1 (5) 4 2 1 12 6 3 20 10 5 12 14 7 5 z z z z z z z z z z z z z                 3 2 2 12 20 6 10 3 5 z z z z z       3 2 12 14 7 5 z z z     26.   3 2 (2 3) 4 3 k k k k        3 2 3 2 2 4 3 3 4 3 k k k k k k k       4 3 2 3 2 8 6 2 12 9 3 k k k k k k       4 3 2 8 6 7 3 k k k k     27. (6 k – 1)(2 k + 3) (6 )(2 3) ( 1)(2 3) k k k      2 12 18 2 3 k k k     2 12 16 3 k k    28. (8 r + 3)( r – 1) Use FOIL. 2 8 8 3 3 r r r     2 8 5 3 r r    29. (3 y + 5)(2 y +1) Use FOIL. 2 6 3 10 5 y y y     2 6 13 5 y y    30. (5 r – 3 s )(5 r – 4 s ) 2 2 25 20 15 12 r rs rs s     2 2 25 35 12 r rs s    31. (9 k + q )(2 k – q ) 2 2 18 9 2 k kq kq q     2 2 18 7 k kq q    32. (.012 x – .17)(.3 x + .54) = (.012 x )(.3 x ) + (.012 x )(.54) + (–.17)(.3 x ) + (–.17)(.54) 2 .0036 .00648 .051 .0918 x x x     2 .0036 .04452 .0918 x x    33. (6.2 m – 3.4)(.7 m + 1.3) 2 4.34 8.06 2.38 4.42 m m m     2 4.34 5.68 4.42 m m    34. 2 p –3[4 p – (8 p + 1)] = 2 p – 3(4 p – 8 p – 1) = 2 p – 3(– 4 p – 1) = 2 p + 12 p + 3 = 14 p + 3 35. 5 k – [ k + (–3 + 5 k )] = 5 k – [6 k – 3] = 5 k – 6 k + 3 = – k + 3

  9. Page 9
    Read page 9 as text

    6 CHAPTER 1 ALGEBRA AND EQUATIONS 36. 2 (3 1)( 2) (2 5) x x x         2 2 3 5 2 4 20 25 x x x x       2 2 3 5 2 4 20 25 x x x x         2 2 3 4 (5 20 ) (–2 25) x x x x       2 15 27 x x     37. R = 5 (1000 x ) = 5000 x C = 200,000 + 1800 x P = (5000 x ) – (200,000 + 1800 x ) = 3200 x – 200,000 38. R = 8.50(1000 x ) = 8500 x C = 225,000 + 4200 x P = (8500 x ) – (225,000 + 4200 x ) = 4300 x – 225,000 39. R =9.75(1000 x ) = 9750 x 2 2 260, 000 ( 3 3480 325) 3 3480 259, 675 C x x x x          2 2 (9750 ) ( 3 3480 259, 675) 3 6270 259, 675 P x x x x x         40. R = 23.50(1000 x ) = 23,500 x 2 2 145, 000 ( 4.2 3220 425) 4.2 3220 144,575 C x x x x          2 2 (23,500 ) ( 4.2 3220 144,575) 4.2 20, 280 144,575. P x x x x x         41. a. According to the bar graph, the net earnings in 2007 were $673 million. b. Let x = 7. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 7 122.5 7 1104 7 2863 505.47 x x x − + − = − + − = According to the polynomial, the net earnings in 2007 were approximately $505 million. 42. a. According to the bar graph, the net earnings in 2015 were $2757 million. b. Let x = 15. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 15 122.5 15 1104 15 2863 2300.75 x x x − + − = − + − = According to the polynomial, the net earnings in 2015 were approximately $2301 million. 43. a. According to the bar graph, the net earnings in 2012 were $1384 million. b. Let x = 12. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 12 122.5 12 1104 12 2863 1022.12 x x x − + − = − + − = According to the polynomial, the net earnings in 2012 were approximately $1022 million. 44. a. According to the bar graph, the net earnings in 2013 were $8 million. b. Let x = 13. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 13 122.5 13 1104 13 2863 1310.13 x x x − + − = − + − = According to the polynomial, the net earnings in 2013 were approximately $1,310 million. 45. Let x = 17. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 17 122.5 17 1104 17 2863 4035.77 x x x − + − = − + − = According to the polynomial, the net earnings in 2017 will be approximately $4036 million.

  10. Page 10
    Read page 10 as text

    SECTION 1.2 POLYNOMIALS 7 46. Let x = 18. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 18 122.5 18 1104 18 2863 5254.28 x x x − + − = − + − = According to the polynomial, the net earnings in 2018 will be approximately $5254 million. 47. Let x = 19. ( ) ( ) ( ) 3 2 3 2 4.79 122.5 1104 2863 4.79 19 122.5 19 1104 19 2863 6745.11 x x x − + − = − + − = According to the polynomial, the net earnings in 2019 will be approximately $6745 million. 48. The figures for 2013 – 2015 seem high, but plausible. To see how accurate these conclusions are, search Starbucks.com for later annual reports. For exercises 49–52, we use the polynomial 3 2 9.5 401.6 6122 25,598. x x x    49. Let x = 10. 3 2 9.5(10) 401.6(10) 6122(10) 25,598 4962     Thus, the costs were approximately $4962 million in 2010. The statement is false. 50. Let x = 15. 3 2 9.5(15) 401.6(15) 6122(15) 25,598 7934.5     Thus, the costs were approximately $7934.5 million in 2015. The statement is true. 51. Let x = 12. 3 2 9.5(12) 401.6(12) 6122(12) 25,598 6451.6     Let x = 15. 3 2 9.5(15) 401.6(15) 6122(15) 25,598 7934.5     Thus, the costs were $6451.6 million in 2012 and $7934.5 million in 2015. The statement is false. 52. Let x = 11. 3 2 9.5(11) 401.6(11) 6122(11) 25,598 5794.9     Let x = 16. 3 2 9.5(16) 401.6(16) 6122(16) 25,598 8456.4     Thus, the costs were $5794.9 million in 2011 and $8456.4 million in 2016. The statement is true. For exercises 53–58, we use the polynomial 3 2 72.85 2082 16,532 59,357. x x x − + − + 53. Let x = 7. 3 2 3 2 72.85 2082 16,532 59,357 72.85(7) 2082(7) 16,532(7) 59,357 20, 663.45 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2007 was approximately $20,663 million. 54. Let x = 10. 3 2 3 2 72.85 2082 16,532 59,357 72.85(10) 2082(10) 16,532(10) 59,357 29,387 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2010 was $29,387 million. 55. Let x = 12. 3 2 3 2 72.85 2082 16,532 59,357 72.85(12) 2082(12) 16,532(12) 59,357 34,896.2 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2012 was approximately $34,896 million. 56. Let x = 15. 3 2 3 2 72.85 2082 16,532 59,357 72.85(15) 2082(15) 16,532(15) 59,357 33,958.25 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2015 was approximately $33,958 million.

  11. Page 11
    Read page 11 as text

    8 CHAPTER 1 ALGEBRA AND EQUATIONS 57. Let x = 13. 3 2 3 2 72.85 2082 16,532 59,357 72.85(13) 2082(13) 16,532(13) 59,357 36, 247.55 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2013 was approximately $36,248 million. Let x = 9. 3 2 3 2 72.85 2082 16,532 59,357 72.85(9) 2082(9) 16,532(9) 59,357 26,103.35 x x x − + − + = − + − + = Thus, the profit for PepsiCo Inc in 2009 was approximately $26,104 million. Therefore, the profit was higher in 2013. 58. By comparing the answers to problems 55 and 56, the profit was higher in 2012. 59. 2 7.2 5005 230, 000 P x x    . Here is part of the screen capture. For 25,000, the loss will be $100,375; For 60,000, there profit will be $96,220. There is a loss at the beginning because of large fixed costs. When more items are made, these costs become a smaller portion of the total costs. 60. In order for the company to make a profit, 2 7.2 5005 230, 000 0 P x x     Set the profit function equal to 0 and solve for x . 2 8 4450 215, 000 0 x x    By the quadratic formula, x ≈ 45 or x ≈ –601. Since x represents a positive number, x = 45. Therefore, between 40,000 and 45,000 calculators must be sold for the company to make a profit. 61. Let x = 100 (in thousands) 2 7.2(100) 5005(100) 230, 000 342,500    d. The profit for selling 100,000 calculators is $342,500. 62. Let x = 150 (in thousands) 2 7.2(150) 5005(150) 230, 000 682, 750    d. The profit for selling 150,000 calculators is $682,750. Section 1.3 Factoring 1. 2 12 24 12 12 2 12 ( 2) x x x x x x x        2. 5 65 5 (1) 5 (13 ) 5 (1 13 ) y xy y y x y x      3.         3 2 2 2 5 5 1 5 1 r r r r r r r r r r r         4.     3 2 2 2 3 8 (3 ) (8) 3 8 t t t t t t t t t t t        

  12. Page 12
    Read page 12 as text

    SECTION 1.3 FACTORING 9 5. 3 2 6 12 18 z z z     2 6 6 (2 ) 6 (3) z z z z z      2 6 2 3 z z z    6. 3 2 5 55 10 x x x     2 5 5 (11 ) 5 (2) x x x x x      2 5 11 2 x x x    7. 2 3 3(2 1) 7(2 1) y y    2 2 (2 1) (3) (2 1) 7(2 1) y y y       2 (2 1) [3 7(2 1)] y y     2 (2 1) (3 14 7) y y         2 2 (2 1) (14 4) 2 2 1 7 2 y y y y       8. 5 3 (3 7) 4(3 7) x x    3 2 3 (3 7) (3 7) (3 7) (4) x x x      3 2 (3 7) (3 7) 4 x x           3 2 (3 7) 9 42 49 4 x x x        3 2 (3 7) 9 42 45 x x x     9. 4 6 3( 5) ( 5) x x    4 4 2 ( 5) 3 ( 5) ( 5) x x x         2 4 ( 5) 3 5 x x           4 2 ( 5) 3 10 25 x x x        4 2 ( 5) 10 28 x x x     10. 2 4 3( 6) 6( 6) x x                2 2 2 2 2 2 2 2 2 2 2 3( 6) (1) 3( 6) 2( 6) 3( 6) 1 2( 6) 3 6 1 2 12 36 3 6 1 2 24 72 3 6 2 24 73 x x x x x x x x x x x x x x                                    11. 2 5 4 ( 1)( 4) x x x x      12. 2 7 6 ( 1)( 6) u u u u      13. 2 7 12 ( 3)( 4) x x x x      14. 2 8 12 ( 2)( 6) y y y y      15.     2 6 3 2 x x x x      16.     2 4 5 5 1 x x x x      17.     2 2 3 3 1 x x x x      18.     2 12 4 3 y y y y      19.     2 3 4 1 4 x x x x      20.     2 2 8 2 4 u u u u      21.     2 9 14 2 7 z z z z      22.     2 6 16 2 8 w w w w      23. 2 10 24 ( 4)( 6) z z z z      24. 2 16 60 ( 6)( 10) r r r r     

  13. Page 13
    Read page 13 as text

    10 CHAPTER 1 ALGEBRA AND EQUATIONS 25. 2 2 9 4 (2 1)( 4) x x x x      26. 2 3 8 4 (3 2)( 2) w w w w      27. 2 15 23 4 (3 4)(5 1) p p p p      28. 2 8 14 3 (4 1)(2 3) x x x x      29. 2 4 16 15 (2 5)(2 3) z z z z      30. 2 12 29 15 (3 5)(4 3) y y y y      31. 2 6 5 4 (2 1)(3 4) x x x x      32. 2 12 1 (4 1)(3 1) z z z z      33. 2 10 21 10 (5 2)(2 5) y y y y      34. 2 15 4 4 (5 2)(3 2) u u u u      35. 2 6 5 4 (2 1)(3 4) x x x x      36. 2 12 7 10 (3 2)(4 5) y y y y      37.    2 3 2 5 3 5 1 a a a a      38.   2 2 6 48 120 6 8 20 6( 10)( 2) a a a a a a         39. 2 2 2 81 (9) ( 9)( 9) x x x x       40. 2 2 17 72 ( 8 )( 9 ) x xy y x y x y      . 41. 2 2 2 2 9 12 4 (3 ) 2(3 )(2) 2 (3 2) p p p p p        42. 2 3 2 (3 2)( 1) r r r r      . 43. 2 2 3 10 ( 2 )( 5 ) r rt t r t r t      . 44. 2 2 2 6 (2 3 )( 2 ). a ab b a b a b      45. 2 2 2 2 2 8 16 ( ) 2( )(4 ) (4 ) ( 4 ) m mn n m m n n m n        46.   2 2 8 16 10 2 4 8 5 2(2 1)(2 5) k k k k k k         47.   2 2 4 12 9 2 3 u u u     48.      2 2 2 9 16 3 4 3 4 3 4 p p p p       49. 2 25 10 4 p p   This polynomial cannot be factored 50.     2 10 17 3 5 1 2 3 x x x x      51.     2 2 4 9 2 3 2 3 r v r v r v     52.     2 2 3 28 7 4 x xy y x y x y      53.   2 2 2 4 4 2 x xy y x y     54.      2 2 16 12 18 2 8 6 9 2 4 3 2 3 u u u u u u         55. 2 3 13 30 (3 5)( 6) a a a a      . 56. 2 3 2 8 (3 4)( 2) k k k k      57. 2 2 21 13 2 (7 2 )(3 ) m mn n m n m n      58. 2 81 100 (9 10)(9 10) y y y     59. 2 2 4 21 ( 7 )( 3 ) y yz z y z y z      60. 2 49 9 a  This polynomial cannot be factored. 61. 2 121 64 (11 8)(11 8) x x x     62.   2 2 2 2 2 2 2 4 14 49 4 2( )(7 ) (7 ) 4 56 196 4( 7 ) z zy y z z y y z zy y z y               63.   3 3 3 2 64 (4) ( 4) 4 16 a a a a a        64.   3 3 3 2 216 6 ( 6) 6 36 b b b b b       

  14. Page 14
    Read page 14 as text

    SECTION 1.3 FACTORING 11 65.          3 3 3 3 2 2 2 2 8 27 (2 ) (3 ) (2 3 ) 2 2 3 3 (2 3 ) 4 6 9 r s r s r s r r s s r s r rs s                66.   3 3 3 3 2 2 1000 27 (10 ) (3 ) (10 3 ) 100 30 9 p q p q p q p pq q        67. 3 64 125 m  3 3 (4 ) (5) m          2 2 (4 5) 4 4 5 5 m m m           2 (4 5) 16 20 25 m m m     68. 3 216 343 y  3 3 (6 ) (7) y     2 (6 7) 36 42 49 y y y     69. 3 3 1000 y z  3 3 (10 ) ( ) y z   2 2 (10 ) (10 ) (10 )( ) ( ) y z y y z z           2 2 (10 ) 100 10 y z y yz z     70.   3 3 3 3 2 2 125 8 (5 ) (2 ) (5 2 ) 25 10 4 p q p q p q p pq q        71.     4 2 2 2 5 6 2 3 x x x x      72.     4 2 2 2 7 10 2 5 y y y y      73.      4 2 2 2 2 1 1 1 b b b b b b b       74.          4 2 2 2 2 3 4 4 1 2 2 1 z z z z z z z          75.          4 2 2 2 2 12 4 3 2 2 3 x x x x x x x          76.          4 2 2 2 2 4 27 81 4 9 9 2 3 2 3 9 x x x x x x x          77.          4 4 2 2 2 2 2 2 16 81 4 9 4 9 2 3 2 3 4 9 a b a b a b a b a b a b         78.              3 3 6 6 2 2 2 2 4 2 2 4 2 2 2 2 x y x y x y x x y y x y x y x xy y x xy y                79.        3 8 2 2 6 2 2 3 2 2 4 2 8 8 2 2 2 4 x x x x x x x x x x                80.                  9 3 3 6 3 6 6 2 2 3 3 3 3 3 3 3 3 3 2 2 64 64 2 2 2 2 2 2 4 2 2 4 x x x x x x x x x x x x x x x x x x                         81.     4 2 2 2 6 3 3 2 1 3 3 x x x x      is not the correct complete factorization because 2 3 3 x  contains a common factor of 3. This common factor should be factored out as the first step. This will reveal a difference of two squares, which requires further factorization. The correct factorization is        4 2 4 2 2 2 2 6 3 3 3 2 1 3 2 1 1 3 2 1 ( 1)( 1) x x x x x x x x x             82. The sum of two squares can be factored when the terms have a common factor. An example is 2 2 2 2 (3 ) 3 9 9 9( 1) x x x     

  15. Page 15
    Read page 15 as text

    12 CHAPTER 1 ALGEBRA AND EQUATIONS 83. 3 2 2 3 2 2 3 2 ( 2) ( 2)( 2) ( 2)( 4 4) 4 2 8 4 8 6 12 8, x x x x x x x x x x x x x x                   which is not equal to 3 8 x  . The correct factorization is 3 2 8 ( 2)( 2 4). x x x x      84. Factoring and multiplication are inverse operations. If we factor a polynomial and then multiply the factors, we get the original polynomial back. For example, we can factor 2 6 x x   to get ( 3)( 2) x x   . Then if we multiply the factors, we get 2 2 ( 3)( 2) 2 3 6 6 x x x x x x x          Section 1.4 Rational Expressions 1. 2 8 8 56 7 8 7 x x x x x x     2. 3 2 2 27 27 1 81 27 3 3 m m m m m m    3. 2 2 3 2 25 5 5 5 7 35 7 5 p p p p p p     4. 4 2 2 2 2 2 18 6 3 3 4 24 6 4 y y y y y y     5. 5 15 5( 3) 5 4 12 4( 3) 4 m m m m       6. 10 5 5(2 1) 1 20 10 5(2 1) 2 2 z x z x        7. 4( 3) 4 ( 3)( 6) 6 w w w w      8. 6( 2) 6 ( 4)( 2) 4 x x x x        or 6 4 x   9.   2 3 2 2 3 12 3 ( 4) 9 3 3 4 3 y y y y y y y y y      10. 2 2 15 45 15 ( 3) 3 3 9 3 5( 3) 3 3 5( 3) 3 k k k k k k k k k k k k k           11. 2 2 4 4 ( 2)( 2) 2 ( 3)( 2) 3 6 m m m m m m m m m m             12. 2 2 6 ( 3)( 2) 2 ( 4)( 3) 4 12 r r r r r r r r r r             13.        2 2 3 1 2 3 3 1 1 1 1 x x x x x x x x x            14.      2 2 2 2 4 4 2 2 2 2 4 z z z z z z z z           15. 2 2 3 3 3 8 3 8 3 64 16 2 64 2 a a a a a      16. 2 3 2 3 4 4 2 10 2 10 5 9 18 8 8 9 u u u u u u u u      17. 3 3 3 2 7 14 7 66 7 66 3 11 66 11 14 11 14 x x x y x y y y x x x        18. 2 2 6 21 6 2 2 21 7 x y xy x y y y x y x xy     19. 2 15 (2 ) 15 15 5 3 4(2 ) (2 ) 12 12 4 a b a b c a b a b c c c           20. 2 2 4( 2) 3 4 ( 2) 3 3 8( 2) 4 ( 2) 2 2 x w w x w w x w x         

  16. Page 16
    Read page 16 as text

    SECTION 1.4 RATIONAL EXPRESIONS 13 21. 15 3 10 2 15 3 3 6 3 6 10 2 3(5 1) 3 3 2 2 (5 1) 3(5 1) 3 3 3(5 1) 2 2 4 p p p p p p p p                      22. 2 8 3 12 2 8 3 6 3 6 3 12 2( 4) 3 6 3( 4) 6( 4) 6 1 18( 4) 18 3 k k k k k k k k                 23. 9 18 3 6 9( 2) 3( 2) 6 12 15 30 6( 2) 15( 2) 27( 2)( 2) 27 3 90( 2)( 2) 90 10 y y y y y y y y y y y y                   24. 12 24 6 12 12( 2) 6( 2) 36 36 8 8 36( 1) 8( 1) 2 3( 2) 3( 1) 4( 1) 2 4( 1) 4 3( 1) 3( 2) 9 r r r r r r r r r r r r r r r r                         25. 2 2 2 2 4 12 9 4 12 20 2 10 2 10 20 9 4( 3) ( 5)( 4) 2( 5) ( 3)( 3) 4( 3)( 5)( 4) 2( 5)( 3)( 3) 2( 4) 3 a a a a a a a a a a a a a a a a a a a a a a a a                                26.     2 2 2 6 18 12 16 4 12 9 6 24 6( 3) 4(3 4) 2(3 4) 4( 3) 3 3 2 8 3 2 8 2(3 4) 2 (3 4)( 2) 2 r r r r r r r r r r r r r r r r r                        27. 2 2 2 2 6 3 4 12 2 3 ( 3)( 2) ( 4)( 1) ( 4)( 3) ( 3)( 1) ( 3)( 2)( 4)( 1) 2 ( 4)( 3)( 3)( 1) 3 k k k k k k k k k k k k k k k k k k k k k k k k k k                                28. 2 2 2 2 6 9 2 8 7 12 ( 3)( 2) ( 3)( 3) ( 4)( 2) ( 3)( 4) 3 3 3 4 4 4 4 4 3 4 n n n n n n n n n n n n n n n n n n n n n n n n n                                  Answers will vary for exercises 29 and 30. Sample answers are given. 29. To find the least common denominator for two fractions, factor each denominator into prime factors, multiply all unique prime factors raising each factor to the highest frequency it occurred. 30. To add three rational expressions, first factor each denominator completely. Then, find the lowest common denominator and rewrite each expression with that denominator. Next, add the numerators and place over the common denominator. Finally, simplify the resulting expression and write it in lowest terms. 31. The common denominator is 35 z . 2 1 2 5 1 7 10 7 3 7 5 7 5 5 7 35 35 35 z z z z z z z           32. The common denominator is 12 z . 4 5 4 4 5 3 16 15 1 3 4 3 4 4 3 12 12 12 z z z z z z z           33. 2 2 ( 2) ( 2) 3 3 3 2 2 4 3 3 r r r r r r             34. 3 1 3 1 (3 1) (3 1) 2 1 8 8 8 8 4 y y y y            35. The common denominator is 5 x . 4 1 4 5 1 20 20 5 5 5 5 5 5 x x x x x x x x x           

787 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Mathematics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.