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Solving Polynomial Equations

±2, ± i 7. 5 ± 5 i 3 4. –5,. Solving Polynomial Equations. ALGEBRA 2 LESSON 6-4. Factor each expression. 2. 216 x 3 – 1 3. 8 x 3 + 125 4. x 4 – 5 x 2 + 4 Solve each equation. 5. x 3 + 125 = 0 6. x 4 + 3 x 2 – 28 = 0. (6 x – 1)(36 x 2 + 6 x + 1).

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Solving Polynomial Equations

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  1. ±2, ±i 7 5 ± 5i 3 4 –5, Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor each expression. 2. 216x3 – 1 3. 8x3+ 125 4.x4 – 5x2 + 4 Solve each equation. 5.x3 + 125 = 0 6.x4 + 3x2 – 28 = 0 (6x – 1)(36x2 + 6x + 1) (2x + 5)(4x2 – 10x + 25) (x + 1)(x – 1)(x + 2)(x – 2) 6-4

  2. Assignment 56 • Page 324 12-32 even, 41-52, 71, 76,77

  3. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 pages 324–326  Exercises 1. –2, 1, 5 2. –1, 0, 3 3. 0, 1 4. 0, 8 5. 0, –1, –2 6. 0, –3.5, 1 7. 0, –0.5, 1.5 8. –0.5, 0, 3 9. 1, 7 10. 4.8% 11. about 5.78 ft  6.78 ft  1.78 ft 12. (x + 4)(x2 – 4x + 16) 13. (x – 10)(x2 + 10x + 100) 14. (5x – 3)(25x2 + 15x + 9) 15. 3, 16. –4, 2 ± 2i 3 17. 5, 18. –1, 19. , 20. – , 21. (x2 – 7)(x – 1)(x + 1) 22. (x2 + 10)(x2 – 2) –3 ± 3i 3 2 –5 ± 5i 3 2 1 ± i 3 2 –1 ± i 3 4 1 2 1 2 1 ± i 3 4 6-4

  4. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 23. (x2 – 3)(x – 2)(x + 2) 24. (x – 2)(x + 2)(x – 1)(x + 1) 25. (x – 1)(x + 1)(x2 + 1) 26. 2(2x2 – 1)(x + 1)(x – 1) 27. ±3, ±1 28. ±2 29. ±4, ±2i 30. ±3i, ± 2 31. ± 2, ±i 6 32. ±i 5, ±i 3 33. –1, 3.24, –1.24 34. –9, 0 35. –2, –3, 1, 2 36. 1.71, 0.83 37. 0, 1.54, 8.46 38. 0, 1.27, 4.73 39. –1.04, 0, 6.04 40. (n – 1)(n)(n + 1) = 210; 5, 6, 7 41. about 3.58 cm, about 2.83 cm 42. – , 43. , 44. ±2 2, ±2i 2 45. ±5, ±i 2 3 ± 3i 3 5 6 5 4 3 –2 ± 2i 3 3 6-4

  5. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 46. ±3i, ±i 3 47. 0, ±2, ±1 48. ± 10, ±i 10 49. 0, ± 50. 4, –2 ± 2i 3 51. 0, 3 ± 3 52. – , 0, 4 53. –1, 1, ±i 5 54. –3, –2, 2 55. –1, 3, 3 56. 0, 1, 3 57. 0, 0, 1, 6 58. ± , ±i 59. ± 2, ±i 60. Check students’ work. 61.V = x2(4x – 2), 4 in. by 4 in. by 16 in. 62.x = length, V = x(x – 1)(x – 2), 5 meters 63. – , 1; y = (2x + 5)(x – 1) 64. ±3, ±1; y = (x – 1)(x + 1)(x – 3)(x + 3) 65. –1, 2, 2; y = (x + 1)(x – 2)2 66. –2, 1, 3; y = (x + 2)(x – 1)(x – 3) 67. –4, –1, 3; y = (x + 4)(x + 1)(x – 3) 68. A cubic can only have 3 zeros. 3 2 265 10 1 2 5 2 3 2 6-4

  6. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 76.x2 – 3x – 10 77. 2x2 + x – 3, R 2 71.a. 10 b. 8 and 12 6-4

  7. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x3 – 125. x3 – 125 = (x)3 – (5)3Rewrite the expression as the difference of cubes. = (x – 5)(x2 + 5x + (5)2) Factor. = (x – 5)(x2 + 5x + 25) Simplify. 6-4

  8. 8x3 + 125 = (2x)3 + (5)3Rewrite the expression as the difference of cubes. = (2x + 5)((2x)2 – 10x + (5)2) Factor. = (2x + 5)(4x2 – 10x + 25) Simplify. 5 2 Since 2x + 5 is a factor, x = – is a root. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve 8x3 + 125 = 0. Find all complex roots. The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0. 6-4

  9. –b ± b2 – 4ac 2a x = Quadratic Formula =    Substitute 4 for a, –10 for b, and 25 for c. –(–10) ± (–10)2 – 4(4)(25) 2(4) – (–10) ± –300 8 = Use the Order of Operations. 10 ± 10i 3 8 = –1 = 1 = Simplify. 5 2 5 ± 5i 3 4 5 ± 5i 3 4 The solutions are – and . Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) 6-4

  10. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Factor x4 – 6x2 – 27. Step 1:  Since x4 – 6x2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables. x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27 Rewrite in the form of a quadratic expression. = a2 – 6a – 27 Substitute a for x2. Step 2:  Factor a2 – 6a – 27. a2 – 6a – 27 = (a + 3)(a – 9) Step 3:  Substitute back to the original variables. (a + 3)(a – 9) = (x2 + 3)(x2 – 9) Substitute x2 for a. = (x2 + 3)(x + 3)(x – 3) Factor completely. The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3). 6-4

  11. x = ± 3 or x = ± i 5 Solve for x, and simplify. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solve x4 – 4x2 – 45 = 0. x4 – 4x2 – 45 = 0 (x2)2 – 4(x2) – 45 = 0    Write in the form of a quadratic expression. Think of the expression as a2 – 4a – 45, which factors as (a – 9)(a + 5). (x2 – 9)(x2 + 5) = 0 (x – 3)(x + 3)(x2 + 5) = 0 x = 3 or x = –3 or x2 = –5 Use the factor theorem. 6-4

  12. Assignment 56 • Page 324 13-31 odd, 55,57,59, 78

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