html5-img
1 / 12

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).

gladys
Download Presentation

Solving Polynomial Equations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related