3-67 (Every fourth, and also 61, 65, 69; pp.309-310)

SECTION 3.3 . Dividing Polynomials 309 EXAMPLE 5 = Factoring a Polynomial Using the Factor Theorem 1/10 -7 6 Let P(x) = x - 7x + 6. Show that P( 1) = 0, and use this fact to factor P(x) 1 1 6 completely. 160 SOLUTION Substituting, we see that P( 1) = 1' - 7 . 1 + 6 = 0. By the Factor Theorem this means that x - 1 is a factor of P(x). Using synthetic or long division (shown in the margin), we see that x - 12 + 0r - 7: + 6 P(x ) - 13- 7*+6 Given polynomial - (x - 1)(x2 +x -6) See margin X- 7x - (x - 1)(x - 2)(x +3) Factor quadratic x3 + x - 6 - 68 + 0 . Now Try Exercises 53 and 57 Ox + 6 EXAMPLE 6 = Finding a Polynomial with Specified Zeros Find a polynomial of degree four that has zeros -3, 0, 1, and 5, and the coefficient of x is -6. SOLUTION By the Factor Theorem, x - (-3), x - 0, x - 1, and x - 5 must all be factors of the desired polynomial. Let 20- P(x) = (x+3)(x - 0)(x - 1)(x-5) =x4 - 3x3 - 13x2 + 15x The polynomial P(x) is of degree 4 with the desired zeros, but the coefficient of x3 is -3, not -6. Multiplication by a nonzero constant does not change the degree, so the desired polynomial is a constant multiple of P(x) . If we multiply P(x) by the con- stant 2, we get e(x) = 2x4 - 6x3 - 26x2 + 30x FIGURE 1 ((x) = 2x(x + 3)(x - 1)(x-5) which is a polynomial with all the desired properties. The polynomial Q is graphed in has zeros -3, 0, 1, and 5, and the Figure 1. Note that the zeros of Q correspond to the x-intercepts of the graph. coefficient of x is -6 Now Try Exercises 63 and 67 3.3 EXERCISES CONCEPTS SKILLS 1. If we divide the polynomial P by the factor x - c and we 3-8 Division of Polynomials Two polynomials P and D are obtain the equation P(x) = (x - c)@(x) + R(x), then we say given. Use either synthetic or long division to divide P(x) by that x - c is the divisor, Q(x) is the , and R(x) is D(x), and express the quotient P(x) /D(x) in the form the P(x ) - e ( x ) + - R(x ) D ( x ) D(x) 2. (a) If we divide the polynomial P(x) by the factor x - c and we obtain a remainder of 0, then we know that c is a !(3.) P (x ) = 2x2 - 5x -7, D(x ) = x - 2 4. P(x ) = 3x3 + 9x2 - 5x - 1, D(x ) = *+4 of P 5. P(x) = 4x2 - 3x - 7, D(x) = 2x - 1 (b) If we divide the polynomial P(x) by the factor x - c and we obtain a remainder of k, then we know that 6. P(x) = 6x3 + x2 - 12x + 5, D(x) = 3x- 4 P ( C ) = 7 . P ( x ) = 2 x 4 - x 3 + 9x ? , D ( x ) = 1 2 + 4 8. P(x) = 2x5 + x3 - 2x2 + 3x - 5, D(x) = x2 - 3x+ 1310 CHAPTER 3 . Polynomial and Rational Functions 9-14 = Division of Polynomials Two polynomials P and D are 42. P(x) -x-+x+5. cm-1 given. Use either synthetic or long division to divide P(x) by D(x), and express P in the form 43 P(x ) - 1+ 2x3 -7. cm-2 P(x) - D(x) . (x) + R(x ) 44. P(.x ) - 2.13-21.13+9x -200. c = 11 *. 9. P(x) - -x- 2+ 6. D(x) -x+ 1 45. P(.x) - 5x* + 30' - 40.x3 + 36r + 14. cm-7 10. P(x) = x' + 2 - 10x, D(x) =x -3 46. P(x) - 68' + 108+ + + 1. c=-2 11, P(.x) = 213 - 313- 2x, D(x) = 2x -3 47. P(x) - x - 30-1. e=3 12 . P(x) = 45 + 7x +9. D(x) = 2x + 1 48. P(.x) - -210 + 7x3 + 4014 - 7.x2 + 10x + 112. c = -3 13. P(x) = 8x4 + 4x + 68, D(x) = 20" + 1 49. P(x) = 3x3 + 413 - 2.* + 1, c=] 14. P(x) = 27x3 - 9x4 + 3x2 - 3. D(x) = 3.12 - 30 + 1 50. P(x) = 1- x+1. 6=1 51, P(x) = 3 + 2x2- 3x - 8, c = 0.1 15-24 = Long Division of Polynomials Find the quotient and remainder using long division. 52. Remainder Theorem Let x? - 3x + 7 P(x) = 6x7 - 40x6 + 16.x5 - 200x4 13 x - 2 *+ 3 - 60x3 - 69x2 + 13x - 139 4x3 + 2x2-2x -3 18. * + 3x2 + 4x + 3 Calculate P(7) by (a) using synthetic division and (b) substi 2.x + 1 3.x + 6 tuting x = 7 into the polynomial and evaluating directly. X19. x + 2x + 1 20. - 3x3 + x - 2 53-56 - Factor Theorem Use the Factor Theorem to show that 1 - x+ 3 x2 - 5x + 1 x - c is a factor of P(x) for the given value(s) of c. 21. Ox + 2x2 + 22x 9.x2 - x + 5 53. P(x) = x3 - 3x2 + 3x - 1, c = 1 2x2 + 5 22. 3x2 - 7x 54. P(x) = x3 + 2x2 - 3x - 10, c = 2 23 *+x4+x 2+1 x 2 + 1 24. 2x5 - 7x4 - 13 55) P ( x ) = 2x 3 + 7 x 2 + 6x - 5 , c = 2 4x 2 - 6x + 8 56. P(x) = x4+ 3x3 - 16x2 - 27x + 63, c= 3, -3 25-38 - Synthetic Division of Polynomials Find the quotient and remainder using synthetic division. 57-62 Factor Theorem Show that the given value(s) of c are zeros of P(x), and find all other zeros of P(x). 25. 2x2 - 5x + 3 x - 3 26. - * + x - 4 x + 1 .57. P( x) = x3 + 2x2 - 9x - 18, c = -2 58 . P (x ) = x3 - 5x2 - 2x + 10, c = 5 27. 3x + x x + 1 28. 4x2 - 3 x - 2 59. ) P ( x ) = x3 - x2 - 11x + 15, c = 3 29. X *+ 2x2 + 2 x + 1 3x - 12x2 - 9x + 1 60. P(x) = 3x4 - x3 - 21x2 - 11x + 6, c = -2, 3 * + 2 - 5 61. P(x) = 3x4 - 8x3 - 14x2 + 31x + 6, C = - 2, 3 * - 8x + 2 32. *4 - x3 + x2 - x + 2 62. P(x) = 2x4 - 13x3 + 7x2 + 37x + 15, c = -1, 3 * + 3 x - 2 33. x + 3x3 - 6 3 - 9x2 + 27x - 27 63-66 - Finding a Polynomial with Specified Zeros Find a x - 1 34. x - 3 polynomial of the specified degree that has the given zeros. 63. Degree 3; zeros - 1, 1, 3 35 2x3 + 3x2 - 2x + 1 x - 2 64. Degree 4; zeros -2, 0, 2, 4 36. 6x4 + 10x3+ 5x2 + x + 1 65. Degree 4; zeros - 1, 1, 3, 5 66. Degree 5; zeros -2, -1, 0, 1, 2 37. x5 - 27 x - 3 38. x* - 16 67-70 - Polynomials with Specified Zeros Find a polynomial * +2 of the specified degree that satisfies the given conditions. 39-51 - Remainder Theorem Use synthetic division and the .(7, Degree 4; zeros - 2, 0, 1, 3; coefficient of x' is 4 Remainder Theorem to evaluate P(c). *(39) P( x ) = 4x2 + 12x + 5 , C=- 1 68. Degree 4; zeros - 1, 0, 2, 2; coefficient of x3 is 3 40. P(x ) = 2x2 + 9x+ 1, c =? 69, Degree 4; zeros -1, 1, V2; integer coefficients and constant term 6 41. P( x ) = x3 + 3x2 - 7x+ 6, c = 2 70. Degree 5; zeros -2, -1, 2, V5; integer coefficients and constant term 40