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Chapter 5, continued 4. G; Let x 5 height. Lesson 5.6 V 5 *wh 5.6 Guided Practice (pp. 371–373) 540 5 (x 1 15)(x 2 3)(x) 1. Factors of the constant term: 61, 63, 65, 615 0 5 x3 1 12x2 2 45x 2 540 Factors of the leading coefficient: 61 0 5 x2(x 1 12) 2 45(x 1 12) 0 5 (x2 2 45)(x 1 12) 3 1 5 15 Possible rational zeros: 6}1, 6}1, 6}1, 6} 1 } The only positive solution is x 5 Ï45 , or x ø 6.7, so the height of the box is about 6.7 inches. 5. A; Simplified list: 61, 63, 65, 615 2. Factors of the constant term: 61, 62, 63, 66 Factors of the leading coefficient: 61, 62 Profit Price per (millions 5 camera of dollars) (dollars) Number + of cameras (millions) Cost per 2 camera (dollars) Number + of cameras (millions) 1 2 3 6 3 2 6 6}2, 6}2, 6}2 1 P 5 (100 2 10x2)(x) 2 30x P 5 100x 2 10x3 2 30x 3 6 9 18 1 2 3. Possible rational zeros: 6}, 6}, 6}, 6}, 6}, 6} 1 1 1 1 1 1 Test x 5 1: 1 When P 5 60: 1 1 24 215 18 1 23 218 23 218 0 Because 1 is a zero of f, f (x) can be writen as: 0 5 10x 3 2 70x 1 60 You know that 2 is a solution, so x 2 2 is a factor of 10x 3 2 70x 1 60. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 2 10 10 3 Simplified list: 61, 62, 63, 66, 6}2 , 6}2 P 5 70x 2 10x 3 60 5 70x 2 10x 3 1 Possible rational zeros: 6}1, 6}1, 6}1, 6}1, 6}2, 0 270 60 20 40 260 20 230 0 So, (x 2 2)(10x2 1 20x 2 30) 5 0. use factoring to find f(x) 5 (x 2 1)(x2 2 3x 2 18) 5 (x 2 1)(x 2 6)(x 1 3) The zeros are 23, 1 and 6. 4. Possible rational zeros: 61, 62, 67, 614 Test x 5 1: 1 1 28 5 14 1 27 22 22 212 28 5 14 21 9 214 29 14 0 that x 5 1 and x 5 23 are the other solutions. 1 27 1 is not a zero. The company could produce 1 million cameras and make the same profit. Test x 5 21: 6. In 2001, t 5 6. 21 1 C 5 20.027(6)4 1 0.32(6)3 2 0.25(6)2 2 4.9(6) 1 51 5 234.992 1 69.12 2 9 2 29.4 1 51 ø 46.73 The average monthly cell phone bill in 2001 was about $46.73. 1 7. V 5 } s2h 3 1 Because 21 is a zero of f, f(x) can be written as: f (x) 5 (x 1 1)(x2 2 9x 1 14) 5 (x 1 1)(x 2 7)(x 2 2) The zeros of f are 21, 2 and 7. 1 48 5 }3 (3x 2 6)2(x) 1 48 5 }3 x(9x2 2 36x 1 36) 0 5 3x 3 2 12x 2 1 12x 2 48 0 5 3x 2(x 2 4) 1 12(x 2 4) 0 5 (3x 2 1 12)(x 2 4) The only positive solution is x 5 4, so the height of the sculpture is 4 feet. Algebra 2 Worked-Out Solution Key 283 Chapter 5, continued 3 1 1 1 5. Possible rational zeros: 61, 63, 6}, 6}, 6}, 6}, 2 2 3 4 3 3 3 1 1 1 1 1 1 } } } } } } } } } 6 4, 66 , 68, 68 , 612, 6 16, 616, 624, 648 3 2 } 2 2 1 220 21 3 6 221 4 214 0 f(x) 5 (x 1 2) + g (x) 5 (x 1 2)1 x 2 }2 2(2x2 1 4x 2 14) 3 2x2 1 4x 2 14 5 0 3 3 1 }} Reasonable values: x 5 2}4, x 5 } , x 5 }2 16 24 6 Ï(4)2 2 4(2)(214) 2(2) 3 Check: x 5 2}4 : 4 220 3 236 24 23 232 4 0 48 48 } 5 21 6 2Ï2 7. } 1 V 5 }3 (x2)(x 2 1) 1 6 5 }3 x2(x 2 1) 3 2}4 is a zero. 1 } 3 The real zeros of f are: 22, }2, 21 1 2Ï2 , and 21 2 2Ï2 . 18 5 x3 2 x2 2 3 f (x) 5 x 1 }4 (48x2 2 32x 1 4) 0 5 x3 2 x2 2 18 Possible rational zeros: 61, 62, 63, 66, 69, 618 5 1 x 1 }4 2(4)(12x2 2 8x 1 1) 3 2 21 0 218 2 2 4 1 1 2 214 1 21 0 218 3 6 18 1 5 (4x 1 3)(6x 2 1)(2x 2 1) 3 1 4 6 1 2 The real zeros of f are 2}, }, and }. 6. Possible rational zeros: 61, 62, 63, 66, 67, 614, 3 7 1 21 621, 642, 6}2 , 6}2, 6}2, 6} 2 3 1 2 6 0 The other solutions, which satisfy x 2 1 2x 1 6, are } x 5 21 6 iÏ 5 and can be discarded because they are imaginary. The only real solution is x 5 3. The base is 3 3 3 feet. The height is x 2 1 or 2 feet. 7 3 Reasonable zeros: x 5 2}2 , x 5 22, x 5 }2 , x 5 2 7 2}2 22 218 219 27 7 } 2 22 2 5 2 2 5 42 2273 4 2105 } 4 } 211 77 2 39 } 2 218 219 42 22 40 242 1 220 21 0 f (x) 5 (x 1 2)(2x 1 x 2 20x 1 21) 5 (x 1 2) + g(x) 2 1 Possible rational zeros of g(x): 61, 63, 67, 621, 6}2, 3 7 21 6}2, 6}2 , 6} 2 3 Reasonable zeros: x 5 }2 , x 5 2 284 Algebra 2 Worked-Out Solution Key Skill Practice 1. If a polynomial function has integer coefficients, then p every rational zero of the function has the form }q where p is a factor of the constant term and q is a factor of the leading coefficient. 2. To shorten the list, draw the graph to see approximately 24 3 5.6 Exercises (pp. 374–377) where the zeros are. 3. f(x) 5 x3 2 3x 1 28 Factors of constant term: 61, 62, 64, 67, 614, 628 Factors of leading coefficient: 61 1 2 4 7 14 28 , 6} Possible rational zeros: 6}1, 6}1, 6}1, 6}1 , 6} 1 1 Simplified list: 61, 62, 64, 67, 614, 628 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 3 2}4 } 24 6 Ï128 4 x 5 }} 5 } Chapter 5, continued Simplified list: 61, 63, 65, 615, 6}2 , 6}2 , Factors of constant term: 61, 62, 65, 610 5 1 1 15 3 1 5 10. h(x) 5 6x 3 2 3x 2 1 12 Simplified list: 61, 62, 65, 610 Factors of the constant term: 61, 62, 63, 64, 66, 612 5. f(x) 5 2x4 1 6x3 2 7x 1 9 Factors of the leading coefficient: 61, 62, 63, 66 Factors of constant term: 61, 63, 69 1 9 3 1 9 12 Possible rational zeros: 6}1 , 6}1, 6}1, 6}2 , 6}2, 6}2 3 1 3 2 4 6 Possible rational zeros: 6}1, 6}1, 6}1, 6}1, 6}1 , Factors of leading coefficient: 61, 62 3 5 6} 8 10 1 Possible rational zeros: 6}, 6}, 6}, 6} 1 3 1 15 5 1 2 1 15 6}2 , 6} , 6}4 , 6}4, 6}4, 6} , 6}8, 6}8 , 6}8 , 7 2 Factors of leading coefficient: 61 2 1 3 6 4 12 , 6}2 , 6} , 6}2, 6}2 , 6}2 , 6} , 6} 2 1 2 9 Simplified list: 61, 63, 69, 6}2, 6}2, 6}2 1 2 3 4 6 12 6 4 12 1 2 3 6}3 , 6}3, 6}3, 6}3 , 6}3, 6} , 6}6, 6}6, 6}6, 3 6. h(x) 5 2x 3 1 x 2 2 x 2 18 Factors of constant term: 61, 62, 63, 66, 69, 618 6}6 , 6}6, 6} 6 Factors of leading coefficient: 61, 62 Simplified list: 61, 6 2, 63, 64, 66, 612, 3 6 9 18 1 2 1 , 6}2, Possible rational zeros: 6}1, 6}1 , 6}1, 6}1 , 6}1 , 6} 1 3 6 9 18 2 6}2 , 6}2 , 6}2, 6}2 , 6} 2 3 9 1 Simplified list: 61, 62, 63, 66, 69, 618, 6}2, 6}2, 6}2 6}2 , 6}2, 6}3, 6} , 6}3 , 6}6 3 7. g (x) 5 4x5 1 3x3 2 2x 2 14 3 1 7 2 14 1 624 1 212 35 224 1 211 24 24 0 14 1 2 7 14 6} , 6}4, 6}4 , 6}4, 6} 2 4 7 1 1 8. f(x) 5 3x4 1 5x3 2 3x 1 42 Factors of the constant term: 61, 62, 63, 66, 67, 614, 621, 642 3 2 6 7 Possible rational zeros: 6 }1, 6}1, 6}1, 6}1 , 6}1, 14 21 42 1 3 2 6 7 21 42 , 6} 6} 3 3 Simplified list: 61, 62, 63, 66, 67, 614, 2 7 Real zeros are 1, 3, and 8. 12. Possible rational zeros: 61, 62, 64, 67, 68, 614, 628, 656 25 222 56 1 24 226 1 24 226 Test x 5 21: 30 21 1 1 14 , 6} , 6} , 6}3 , 6}3 , 6}3, 6}3 , 6}3 , 6} , 6} 3 1 1 1 1 5 (x 2 1)(x 2 3)(x 2 8) Test x 5 1: Factors of the leading coefficient: 61, 63 1 f(x) 5 (x 2 1)(x 2 11x 1 24) 2 7 Simplified list: 61, 6 2, 6 7, 6 14, 6}2 , 6}2, 6}4, 6}4 211 1 7 2 1 14 621, 642, 6}3 , 6}3, 6}3, 6} 3 1 9. h(x) 5 8x4 1 4x3 2 10x 1 15 25 222 56 21 6 16 26 216 72 Factors of the constant term: 61, 63, 65, 615 Test x 5 2: Factors of the leading coefficient: 61, 62, 64, 68 2 1 3 5 1 Test x 5 1: Factors of the leading coefficient: 61, 62, 64 1 4 11. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612, 1 Possible rational zeros: 6}1, 6}1 , 6}1, 6} , 6}2, 6}2, 6}2, 1 2 1 Factors of the constant term: 61, 62, 67, 614 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 3 1 4. g (x) 5 x 3 2 4x 2 1 x 2 10 15 1 1 , 6}2, Possible rational zeros: 6}1 , 6}1 , 6}1, 6} 1 3 5 15 3 5 15 3 1 1 6}2 , 6}2, 6} , 6}4 , 6}4 , 6}4, 6} , 6}8, 6}8, 2 4 5 15 6}8 , 6} 8 1 25 222 56 2 26 256 23 228 0 f(x) 5 (x 2 2)(x2 2 3x 2 28) 5 (x 2 2)(x 2 7)(x 1 4) Real zeros are 2, 7, and 24. Algebra 2 Worked-Out Solution Key 285 Chapter 5, continued 13. Possible rational zeros: 61, 62, 63, 65, 66, 610, 615, 630 Test x 5 1: 1 1 0 231 230 1 1 230 1 230 260 1 1 231 230 21 1 30 21 230 0 5 (x 1 1)(x 2 6)(x 1 5) Real zeros are 21, 6, and 25. 14. Possible rational zeros: 61, 62, 63, 64, 66, 68, 69, 612, 618, 624, 636, 672 21 1 44 24 1 8 34 78 8 34 78 102 1 8 29 272 1 9 0 9 2 1 29 272 21 27 16 7 216 256 1 1 1 8 29 272 2 20 22 10 11 250 20 24 0 6 20 24 2 16 52 8 26 76 1 6 20 24 22 28 224 4 12 0 h(x) 5 (x 1 1)(x 1 2)(x2 1 4x 1 12) 0 5 x2 1 4x 1 12 }} 24 6 Ï(4)2 2 4(1)(12) x 5 }} 2(1) 16. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612, 624 272 22 212 42 6 221 230 1 1 1 22 29 10 224 1 21 210 0 21 210 0 224 22 29 10 224 21 3 6 216 23 26 16 240 Test x 5 21: 29 272 3 33 72 11 24 0 h(x) 5 (x 2 3)(x2 1 11x 1 24) 5 (x 2 3)(x 1 8)(x 1 3) The real zeros are 3, 28, and 23. 286 6 Test x 5 1: Test x 5 3: 3 1 8 1 224 The only real solutions are 21 and 22. 29 1 220 } 8 1 26 5 22 6 2iÏ 2 , which are not real. Test x 5 22: 22 21 1 Test x 5 2: 2 24 Test: x 5 22: 22 8 1 44 Test: x 5 2: Test x 5 21: 21 26 Possible rational zeros of g (x): 61, 62, 63, 64, 66, 68, 612, 624 1 272 0 7 1 Test x 5 1: 1 26 h(x) 5 (x 1 1)(x3 1 6x2 1 20x 1 24) 5 (x 1 1) + g (x) g(x) 5 (x 1 1)(x2 2 x 2 30) 1 7 Test x 5 21: 0 1 1 1 Test x 5 21: 21 624 Test x 5 1: Algebra 2 Worked-Out Solution Key 21 1 1 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 1 15. Possible rational zeros: 61, 62, 63, 64, 66, 68, 612, Chapter 5, continued f (x) 5 (x 2 1)(x 1 1)(x2 1 2x 2 8) Test x 5 2: 2 1 22 29 10 224 2 0 218 216 0 29 28 240 1 5 (x 2 1)(x 1 1)(x 1 4)(x 2 2) The real zeros are 1, 21, 24, and 2. 18. Possible rational zeros: 61, 65, 625 Test x 5 1: 1 Test x 5 22: 22 1 1 22 29 10 224 22 8 2 224 24 21 12 248 Test x 5 3: 29 10 224 3 3 218 224 1 1 Test x 5 23: 26 28 248 1 23 1 1 240 225 1 1 215 255 1 215 255 280 0 216 240 225 21 1 15 25 215 225 0 Test x 5 21: 1 21 1 g (x) 5 (x 1 1)(x 2 x 2 15x 2 25) 5 (x 1 1) + f(x) 3 2 Possible zeros of f(x): 61, 65, 625 22 29 10 224 23 15 218 24 25 6 28 0 Test x 5 5: 5 1 21 215 225 5 20 25 4 5 1 f(x) 5 (x 1 3)(x3 2 5x2 1 6x 2 8) 5 (x 1 3) + g(x) 0 Possible rational zeros of g (x): 61, 62, 64, 68 g(x) 5 (x 1 1)(x 2 5)(x 1 4x 1 5) Test x 5 4: x2 1 4x 1 5 5 0 4 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 216 1 21 22 3 0 1 1 1 25 6 28 4 24 8 21 2 0 2 }} 24 6 Ï(4)2 2 4(1)(5) x 5 }} 2(1) 5 22 6 i, which are not real. The only real zeros are 21 and 5. f(x) 5 (x 1 3)(x 2 4)(x2 2 x 1 2) x2 2 x 1 2 5 0 1 1 19. Possible rational zeros: 61, 62, 64, 68, 616, 6}, 6} 2 4 Reasonable rational zeros: x 5 1 } 1 6 Ï1 2 4(1)(2) 2 Test x 5 1: x 5 }} 1 } 1 6 iÏ7 4 5} which are not real. 2 The real solutions are 23 and 4. 17. Possible rational zeros: 61, 62, 64, 68 Test x 5 1: 1 1 1 2 29 22 8 1 3 26 28 3 26 28 0 f(x) 5 (x 2 1)(x 1 3x 2 6x 2 8) 5 (x 2 1) + g (x) 2 Possible rational zeros of g (x): 61, 62, 64, 68 1 16 4 4 216 4 216 0 5 (x 2 1)(4)(x2 1 x 2 4) 0 5 x2 1 x 2 4 }} 21 6 Ï(1) 2 2 4(1)(24) x 5 }} 2(1) } 21 6 Ï 17 5} 2 } } 21 1 Ï17 21 2 Ï17 Test x 5 21: 1 220 f (x) 5 (x 2 1)(4x 2 1 4x 2 16) 3 21 4 0 3 26 28 21 22 8 2 28 0 ,} Real zeros: 1, } 2 2 Algebra 2 Worked-Out Solution Key 287 Chapter 5, continued 3 1 20. Possible rational zeros: 61, 63, 65, 615, 6}, 6}, 2 2 5 15 3 5 15 1 } } } } } } 62, 6 2 , 64, 64, 64, 6 4 3 5 Reasonable zeros: x 5 21, x 5 }2, x 5 }2 Test x 5 21: 21 4 4 212 21 15 24 16 215 216 15 0 28 24 4 8 2 22 24 0 The real zeros are 22, 21, and 2. 5 The real zeros are 21, }2 , and }2 . 3 1 21. Possible rational zeros: 61, 63, 6 5, 615, 6}, 6}, 2 2 5 15 5 5 1 1 6}2, 6} , 6}3, 6}3 , 6}6 , 6}6 2 5 1 Reasonable zeros: x 5 23, x 5 2}3, x 5 }2 Test x 5 23: 3 9 1 25. Possible rational zeros: 61, 63, 69, 6}, 6}, 6} 2 2 2 3 Reasonable zeros: x 5 21, x 5 }2, x 5 3 Test x 5 21: 21 2 2 27 0 9 22 9 29 29 9 0 g(x) 5 (x 1 1)(2x2 2 9x 1 9) 5 (x 1 1)(2x 2 3)(x 2 3) 3 25 16 215 218 221 15 7 25 0 6 28 5 (x 1 2)(2x 1 2)(x 2 2) 5 (x 1 1)(2x 2 3)(2x 2 5) 23 2 2 f(x) 5 (x 1 2)(2x2 2 2x 2 4) f (x) 5 (x 1 1)(4x2 2 16x 1 15) 3 22 6 The real zeros are 21, }2 , and 3. 3 1 26. Possible rational zeros: 61, 63, 65, 6 15, 6 }, 6}, 2 2 5 15 6}2, 6} 2 5 Reasonable zeros: x 5 2}2, x 5 1, and x 5 3 5 (x 1 3)(2x 2 1)(3x 1 5) 5 1 5 Test x 5 2}2: 5 Real zeros are 23, }2, and 2}3. 2}2 23 214 15 25 20 215 2 28 6 0 1 2 2 22. Possible rational zeros: 61, 62, 6 4, 68, 616, 8 16 1 2 4 6 }3, 6 }3 , 6 }3, 6 }3 , 6 } 3 2 Reasonable zeros: x 5 } , x 5 2, x 5 4 3 2 Test x 5 }3 : 2 } 23 20 236 16 3 22 12 216 18 224 0 23 f (x) 5 1 x 2 }3 2(23x2 1 18x 2 24) 2 5 1 x 2 }3 2(3)(2x2 1 6x 2 8) 2 5 (3x 2 2)(2x 1 2)(x 2 4) 2 Real zeros are }3, 2, and 4. 3 9 1 23. C; Possible zeros: 61, 63, 69, 6}, 6}, 6} 2 2 2 1 24. Possible zeros: 61, 62, 64, 68, 6} 2 Reasonable zeros: x 5 22, x 5 21, and x 5 2 Test x 5 22: 288 Algebra 2 Worked-Out Solution Key 5 h(x) 5 x 1 }2 (2x2 2 8x 1 6) 5 1 x 1 }2 2(2)(x2 2 4x 1 3) 5 5 (2x 1 5)(x 2 3)(x 2 1) 5 The real zeros are 2}2 , 3, and 1. 1 27. Possible rational zeros: 61, 62, 63, 64, 66, 612, 6}, 3 2 4 6}3 , 6}3 1 Reasonable zeros: x 5 24, x 5 2}3, and x 5 3 Test x 5 24: 24 3 3 4 235 212 212 32 12 23 0 28 f (x) 5 (x 1 4)(3x 2 8x 2 3) 5 (x 1 4)(3x 1 1)(x 2 3) 2 1 The real zeros are 24, 2}3 , and 3. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. f (x) 5 (x 1 3)(6x 1 7x 2 5) 2 Chapter 5, continued 1 28. Possible real zeros: 61, 62, 63, 64, 66, 612, 6}, 3 2 4 } } 63, 63 2 Reasonable zeros: x 5 26, x 5 21, x 5 }3 Test x 5 26: 26 4 212 218 26 12 22 0 3 1 5 (x 1 4)1 x 2 }2 2(2)(x2 1 x 1 1) 1 5 (x 1 4)(2x 2 1)(x2 1 x 1 1) }} 21 6 Ï(1)2 2 4(1)(1) 2(1) x 5 }} } 21 6 iÏ3 , which are imaginary numbers 5} 2 1 The only real solutions are 24 and }2 . f (x) 5 (x 1 6)(3x 1 x 2 2) 2 5 (x 1 6)(3x 2 2)(x 1 1) 1 31. Possible rational zeros: 61, 62, 64, 6} 2 2 The real zeros are 26, 21, and }3 . Reasonable zeros: x 5 22 and x 5 2 7 1 29. Possible real zeros: 61, 62, 67, 614, 6}, 6} 2 2 7 Reasonable zeros: x 5 2}2 , x 5 21, and x 5 2 7 Test x 5 2}2 : 7 2}2 2 5 211 214 2 1 x2 1 x 1 1 5 0 19 3 g(x) 5 (x 1 4)1 x 2 }2 2 (2x2 1 2x 1 2) 27 7 14 22 24 0 1 Test x 5 22: 22 2 2 2 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 2 5 (2x 1 7)(x 2 2)(x 1 1) 9 5 3 24 28 24 24 4 1 1 21 0 g(x) 5 (x 1 4)(2x 1 x 1 x 2 1) 5 (x 1 4) + f(x) 2 1 Possible real solutions of f(x): 61, 6}2 1 Test x 5 }2 : 1 } 2 1 1 21 2 2 0 3 22 0 25 3 22 4 22 2 21 1 0 }} } 1 6 iÏ 7 The real solutions are x 5 22 and x 5 2. Test x 5 24: 2 25 5} , which are not real. 4 1 Resonable zeros: x 5 24 and x 5 }2 2 4 1 6 Ï(21)2 2 4(2)(1) x 5 }} 2(2) 1 30. Possible real solutions: 61, 62, 64, 6} 2 1 26 2x2 2 x 1 1 5 0 7 1 10 h(x) 5 (x 1 2)(x 2 2)(2x2 2 x 1 1) The real zeros are 2}2 , 2, and 21. 1 24 Test x 5 2: 2 7 3 24 h(x) 5 (x 1 2)(2x3 2 5x2 1 3x 2 2) 5 (x 1 2) + f(x) 5 1 x 1 }2 2(2)(x2 2 x 2 2) 2 4 1 2 2 27 Possible rational zeros of f: 61, 62, 6}2 7 g(x) 5 x 1 }2 (2x2 2 2x 2 4) 24 21 32. Possible rational zeros: 61, 62, 63, 64, 66, 612, 1 2 4 6}3, 6}3, 6}3 Reasonable zeros: x 5 23, and x 5 4 Test x 5 23: 23 3 3 26 232 35 212 29 45 239 12 215 13 24 0 h(x) 5 (x 1 3)(3x3 2 15x2 1 13x 2 4) 5 (x 1 3) + g(x) 1 2 4 Possible rational zeros of g: 61, 62, 64, 6}3 , 6}3, 6}3 Algebra 2 Worked-Out Solution Key 289 Chapter 5, continued Test x 5 4: 3 3 Test x 5 21: 215 13 24 12 212 4 23 1 21 0 1 h(x) 5 (x 1 3)(x 2 4)(3x 2 3x 1 1) 2 }} 3 6 Ï 32 2 4(3)(1) 3 6 iÏ3 x 5 }} 5} , which are not real. 6 2(3) } The only real solutions are 23 and 4. 33. Possible rational zeros: 61, 62, 63, 65, 66, 610, 3 5 0 37 230 24 26 252 30 1 1 213 26 215 0 0 26 11 26 1 25 6 25 6 0 Test x 5 23: 23 15 2 213 26 215 2 211 15 211 15 0 5 23 22 25 3 26 3 0 6 23 21 0 22 1 0 2 h(x) 5 (x 1 3)(2x4 2 x3 2 2x 1 1) 5 (x 1 3) + f(x) 1 Possible rational zeros of f: 61, 6}2 1 Test x 5 }2: 1 2 } 2 5 (x 1 2)(x 2 1)(2x 2 5)(x 2 3) 5 The real zeros are 22, 1, }2 , and 3. 2 34. Possible rational zeros: 61, 62, 63, 64, 66, 612 Reasonable zeros: x 5 22, x 5 21, x 5 1, x 5 2, and x 5 3 Test x 5 22: 1 26 1 f (x) 5 (x 1 2)(x 2 1)(2x2 2 11x 1 15) 1 11 Reasonable zeros: x 5 23, x 5 }2, and x 5 1 Test x 5 1: 22 26 f(x) 5 (x 1 2)(x 1 1)(x 2 1)(x2 2 5x 1 6) 6}2, 6}2, 6} 2 2 6 3 1 35. Possible rational zeros: 61, 63, 6}, 6} 2 2 1 2 211 The real zeros of f are 22, 21, 1, 2, and 3. Possible rational zeros of g: 61, 63, 65, 615, 6}2, 1 6 5 (x 1 2)(x 1 1)(x 2 1)(x 2 2)(x 2 3) 29 f (x) 5 (x 1 2)(2x3 2 13x2 1 26x 2 15) 5 (x 1 2) + g(x) 5 21 Test x 5 1: 1 Test x 5 22: 3 26 5 (x 1 2)(x 1 1) + h(x) 5 2 5 15 Reasonable zeros: x 5 22, x 5 1, x 5 }2 , and x 5 3 2 5 Possible rational zeros of h: 61, 62, 63, 66 615, 630, 6}2 , 6}2, 6}2, 6} . 2 22 25 f (x) 5 (x 1 2)(x 1 1)(x3 2 6x2 1 11x 2 6) 3x2 2 3x 1 1 5 0 1 1 23 25 15 4 212 22 10 210 210 12 25 5 5 26 0 f (x) 5 (x 1 2)(x4 2 5x3 1 5x2 1 5x 2 6) 5 (x 1 2) + g(x) Possible rational zeros of g: 61, 62, 63, 66 21 0 22 1 1 0 0 21 0 0 22 0 1 2 1 h(x) 5 (x 1 3) x 2 }2 (2x3 2 2) 5 (x 1 3)1 x 2 }2 2(2)(x3 2 1) 1 5 (x 1 3)(2x 2 1)(x 2 1)(x2 1 x 1 1) x2 1 x 1 1 5 0 }} 21 6 Ï(1)2 2 4(1)(1) x 5 }} 2(1) } 21 6 iÏ3 5} , which are not real. 2 1 The only real solutions of h are 23, }2, and 1. 36. The error is that only the positive factors were listed as possible rational zeros. Possible zeros: 61, 62, 67, 614 290 Algebra 2 Worked-Out Solution Key Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 4 Chapter 5, continued 37. The error is that the factors of the coefficient were divided by the factors of the constant term. The possible 5 5 5 1 1 1 zeros are: 61, 65, 6}2 , 6}2, 6}3, 6}3, 6}6, 6}6 38. Sample answer: f(x) 5 4x 2 1 3 3 2 1 2 1 4 3 4 h(x) 5 (x 1 2)(x2 2 x 1 1) x2 2 x 1 1 5 0 }} 1 6 Ï(21)2 2 4(1)(1) x 5 }} 2(1) } 1 6 i Ï3 , which are imaginary numbers. 5} 2 Possible zeros: 61, 63, 6}, 6}, 6}, 6} 39. B; y The only real zero is 22, which matches graph A. 44. It is not possible for a cubic function to have more than 3 real zeros. At most a cubic function can have 3 real zeros. The number of possible zeros is determined by degree. The cubic function will either have 1 real zero or 3 real zeros, and never no zeros because the graphs of the functions cross the x-axis at least once. ( ( 4 , 5 20 0 2 x ( ( 3 , 4 (2 , 0( 3 2 0 Problem Solving 40. sr factors. Each factor of a0 (there are s) is divided by each factor of an (there are r), then at most there are sr possible rational zeros. 41. f(x) 5 x 2 2x 2 x 1 2 3 2 h5*14 63 5 *2(* 1 4) 63 5 *3 1 4*2 0 5 *3 1 4*2 2 63 Possible rational zeros: 61, 62 Possible zeros: 1, 3, 7, 9, 21, 63 Test x 5 1: Test * 5 1: 1 1 1 22 21 2 1 21 22 21 22 0 1 1 1 f (x) 5 (x 2 1)(x 2 2 x 2 2) Real zeros are 21, 1, and 2, which matches graph B. 3 1 42. g (x) 5 x 2 3x 1 2 3 2 Possible rational zeros: 61, 62 1 Test x 5 1: 1 1 23 0 2 1 22 22 22 1 22 5 5 5 5 258 4 0 263 3 21 63 7 21 0 Dimensions: 3 in. 3 3 in. 3 7 in. 0 w5d15 * 5 d 1 35 2000 5 (d 1 35)(d 1 5)(d) 2000 5 (d 2 1 40d 1 175)d 0 5 d 3 1 40d 2 1 175d 2 2000 }} Reasonable zero: d 5 5 Test d 5 5: } } } Real zeros: 1, 1 1 Ï3 , and 1 2 Ï 3 , which matches graph C. 5 1 43. h(x) 5 x 3 1 x 2 2 x 1 2 Possible rational zeros: 61, 62 1 40 175 22000 5 225 2000 400 0 45 V 5 (d 2 5)(d 1 45d 1 400) 2 Test x 5 22: 1 1 46. V 5 * + w + d 2 6 Ï (2)2 2 4(1)(22) x 5 }} 2(1) 1 263 * 5 3 is the only real zero. x2 2 2x 2 2 5 0 22 0 V 5 (* 2 3)(*2 1 7* 1 21) g(x) 5 (x 2 1)(x2 2 2x 2 2) 5 1 6 Ï3 4 Test * 5 3: 5 (x 2 1)(x 2 2)(x 1 1) Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 45. V 5 *2h 1 21 2 22 2 22 21 1 0 The only positive real value for d is 5. Dimensions: Depth 5 d 5 5 feet Width 5 d 1 5 5 10 feet Length 5 d 1 35 5 40 feet Algebra 2 Worked-Out Solution Key 291 Chapter 5, 47. continued V 5 x(x 2 1)(x 2 2) Test t 5 8: 24 5 x(x2 2 3x 1 2) 8 12 2264 2028 23924 212,384 96 21344 5472 12,384 2168 684 1548 0 0 5 x 2 3x 1 2x 2 24 3 2 Possible rational solutions: 61, 62, 63, 64, 66, 68, 612, 624 12 1 48. V 5 } (x2(2x 2 5)) 3 1 } 3 5 3 (2x3 2 5x2) t 5 8 is a solution. d. 9 5 2x3 2 5x2 0 5 2x3 2 5x2 2 9 1 3 9 Possible rational solutions: 61, 63, 69, 6}2, 6}2, 6}2 The only reasonable solution is 8 because the other solutions are either negative or imaginary. The year 1998 corresponds to the solution. 49. a. 20,300 5 210t3 1 140t 2 2 20t 1 18,150 0 5 210t 1 140t 2 20t 2 2150 2 b. Possible solutions: 1, 2, 5 c. 1 210 51. 140 220 22150 210 130 110 130 110 22040 3x x l 210 d 21x 1 6 Not drawn to scale Ramp length 5 (21x 1 6) 2 3x 5 18x 1 6 1 2 210 210 5 210 210 140 220 22150 220 240 440 120 220 21710 140 220 22150 250 450 2150 90 430 0 F 1 G 150 5 3x[3x 1 x 1 6x 1 2x] 2 2 50 5 x[9x2 1 3x] 0 5 9x3 1 3x2 2 50 The possible rational solutions: 61, 62, 65, 610, 625, 1 2 5 10 25 50 1 2 5 25 50 6} , 6} 9 9 5 Test x 5 }3: t 5 5 corresponds to 1999. 5 3 } 9 0 5 12t 4 2 264t 3 1 2028t 2 2 3924t 2 12,384 b. Possible solutions: 1, 2, 3, 4, 6, 8, 9 9 c. Test t 5 4: 2264 2028 23924 212,384 48 2864 4656 2928 2216 1164 732 29456 3 0 250 15 30 50 18 30 0 (9x 2 1 18x 1 30)1 x 2 }53 2 5 0 5 12 Test t 5 6: 6 12 12 2264 2028 23924 212,384 72 21152 5256 7992 2192 876 1332 24392 x 5 }3 is the only positive, real solution. right ramp: * 5 12 ft Algebra 2 Worked-Out Solution Key left ramp: * 5 24 ft 5 h 5 }3 ft h 5 }3 ft w 5 5 ft w 5 5 ft 5 5.6 Mixed Review for TAKS 52. B; The sale price is dependent on the original price. 132s8t17 A 53. F; w 5 } 5 } 5 11s3t8 * 12s5t9 The width is 11s 3t 8 units. 292 10 , 6} , 6} , 6}9, 6}9 , 6}9, 6} , 650, 6}3, 6}3, 6}3, 6} 3 3 3 9 50. a. 56,300 5 12t 4 2 264t 3 1 2028t 2 2 3924t 1 43,916 12 * 5 }3 (18x 1 6) 5 12x 1 4 1 V 5 3x }2(x)(6x 1 2) 1 }2(x)(12x 1 4) t 5 5 is an actual solution. 4 2 d 5 }3 (18x 1 6) 5 6x 1 2 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 3