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Polynomial Functions

Polynomial Functions

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Polynomial Functions

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  1. Simplify each expression by combining like terms. 1. 3x + 5x – 7x2. –8xy2 – 2x2y + 5x2y3. –4x + 7x2 + x Find the number of terms in each expression. 4.bh  5. 1 – x  6. 4x3 – x2 – 9 1 2 Polynomial Functions ALGEBRA 2 LESSON 6-1 (For help, go to Lesson 1-2.) 6-1

  2. Polynomial Functions ALGEBRA 2 LESSON 6-1 Solutions 1. 3x + 5x – 7x = (3 + 5 – 7)x = 1x = x 2. –8xy2 – 2x2y + 5x2y = (–2 + 5)x2y – 8xy2 = 3x2y – 8xy2 3. –4x + 7x2 + x = 7x2 + (–4 + 1)x = 7x2 – 3x 4.bh; 1 term 5. 1 – x; 2 terms 6. 4x3 – x2 – 9; 3 terms 1 2 6-1

  3. Polynomial Functions ALGEBRA 2 LESSON 6-1 Write each polynomial in standard form. Then classify it by degree and by number of terms. a. 9 + x3 b.x3 – 2x2 – 3x4 –3x4 + x3 – 2x2 x3 + 9 The term with the largest degree is x3,so the polynomial is degree 3. The term with the largest degree is –3x4, so the polynomial is degree 4. It has two terms. The polynomial is a cubic binomial. It has three terms. The polynomial is a quartic trinomial. 6-1

  4. xy 0 2.8 2 5 4 6 6 5.5 8 4 Linear model Quadratic model Cubic model Polynomial Functions ALGEBRA 2 LESSON 6-1 Using a graphing calculator, determine whether a linear, quadratic, or cubic model best fits the values in the table. Enter the data. Use the LinReg, QuadReg, and CubicReg options of a graphing calculator to find the best-fitting model for each polynomial classification. Graph each model and compare. The quadratic model appears to best fit the given values. 6-1

  5. Number of Employees Year 1975 60 1980 65 1985 70 1990 60 1995 55 2000 64 To estimate the number of employees for 1988, you can use the Table function option of a graphing calculator to find that ƒ(13) 62.72. According to the model, there were about 62 employees in 1988. Polynomial Functions ALGEBRA 2 LESSON 6-1 The table shows data on the number of employees that a small company had from 1975 to 2000. Find a cubic function to model the data. Use it to estimate the number of employees in 1998. Let 0 represent 1975. Enter the data. To find a cubic model, use the CubicReg option of a graphing calculator. Graph the model. The function ƒ(x) = 0.0096x3 – 0.375x2 + 3.541x + 58.96 is an approximate model for the cubic function. 6-1

  6. Polynomial Functions ALGEBRA 2 LESSON 6-1 pages 303–305  Exercises 1. 10x + 5; linear binomial 2. –3x + 5; linear binomial 3. 2m2 + 7m – 3; quadratic binomial 4.x4 – x3 + x; quartic trinomial 5. 2p2 – p; quadratic binomial 6. 3a3 + 5a2 + 1; cubic trinomial 7. –x5; quintic monomial 8. 12x4 + 3; quartic binomial 9. 5x3; cubic monomial 10. –2x3; cubic monomial 11. 5x2 + 4x + 8; quadratic trinomial 12. –x4 + 3x3; quartic binomial 13.y = x3 + 1 14.y = 2x3 – 12 15. y = 1.5x3 + x2 – 2x + 1 16.y = –3x3 – 10x2 + 100 17.a. males: y = –0.002571x2 + 0.2829x + 67.21 females: y = –0.002286x2 + 0.2514x + 74.82 b. males: y = 0.00008333x3 – 0.007571x2 + 0.3545x + 67.11 females: y = 0.00008333x3 – 0.007286x2 + 0.3231x + 74.72 c. The cubic model is a better fit. 6-1

  7. Polynomial Functions ALGEBRA 2 LESSON 6-1 18.y = x3 – 2x2; 4335 19.y = x3 – 10x2; 2023 20.y = –0.5x3 + 10x2; 433.5 21.y = –0.03948x3 + 2.069x2 – 17.93x + 106.9; 206.07 22. y = –0.007990x3 + 0.4297x2 – 6.009x + 43.57; 26.34 23. y = 0.01002x3 – 0.3841x2 + 5.002x + 2.132; 25.39 24. Check students’ work. 25.x3 + 4x; cubic binomial 26. –4a4 + a3 + a2; quartic trinomial 27. 7; constant monomial 28. 6x2; quadratic monomial 29.x4 + 2x3; quartic binomial 30.x5 + x; quintic binomial 31. a.V = 10 r2 b.V = r3 c.V = r 3 + 10 r2 32. Answers may vary. Sample: Cubic functions represent curvature in the data. Because of their turning points they can be unreliable for extrapolation. 33. –c2 + 16; binomial 34. –9d3 – 13; binomial 1 2 2 3 2 3 2 3 6-1

  8. Polynomial Functions ALGEBRA 2 LESSON 6-1 35. 16x2 – x – 5; trinomial 36. 2x3 – 6x + 17; trinomial 37.a + 4b; binomial 38. –12y; monomial 39. 8x2 – 6y; binomial 40. –3a + 2; binomial 41. 2x3 + 9x2 + 5x + 27; polynomial of 4 terms 42. –4x4 – 3x3 + 5x – 54; polynomial of 4 terms 43. 80x3 – 109x2 + 7x – 75; polynomial of 4 terms 44. 2x3 – 2x2 + 8x – 27; polynomial of 4 terms 45. 6a2 + 3ab – 8; trinomial 46. 8x3 + 2x2; binomial 47. 30x3 – 10x2; binomial 48. 2a3 – 5a2 – 2a + 5; polynomial of 4 terms 49.b3 – 6b2 + 9b; trinomial 50. x3 – 6x2 + 12x – 8; polynomial of 4 terms 51. x4 + 2x2 + 1; trinomial 52. 8x3 + 60x2 + 150x + 126; polynomial of 4 terms 53.a3 – a2b – b2a + b3; polynomial of 4 terms 6-1

  9. Polynomial Functions ALGEBRA 2 LESSON 6-1 54.a4 – 4a3 + 6a2 – 4a + 1; polynomial of 5 terms 55. 12s3 + 61s2 + 68s – 21; polynomial of 4 terms 56.x3 + 2x2 – x – 2; trinomial 57. 8c3 – 26c + 12; trinomial 58.s4 – 2t2s2 + t4; trinomial 59.a.y = 0.7166x + 47.61 y = 0.0009365x3 – 0.0744x2 + 2.2929x + 41.4129 y = –0.00004789x4 + 0.004666x3 – 0.1647x2 + 2.9797x + 40.7831 b. The quartic model fits best. c. 72.2  1015 6-1

  10. Polynomial Functions ALGEBRA 2 LESSON 6-1 60. 2.5  108 cm3 61.a. up 4 units b.y = 4x3 is more narrow. c.y = x3 62. B 63. A 64. A 65.[2] If it is in standard form, the degree is the exponent on the first variable. [1] incorrect reason for why it is easier to find in standard form 66. 2 67. 2 68. none 69. (5, –3) 70. 1 –3 1 5 –8 –3 0 –3 71. 2 –3 –6 5 –2 0 72. –2 –1 –2 –1 –3 –3 –4 –4 6-1

  11. x 1 2 3 5 8 y 3 2.2 3 7 5 y = –0.14401x3 + 1.7805x2 – 5.29x + 6.62, y = –0.1319x2 + 1.682x + 0.37 Polynomial Functions ALGEBRA 2 LESSON 6-1 Write each polynomial in standard form. Then classify it by degree and by number of terms. 1. –x2 + 2x + x22. 7x2 + 10 + 4x33. 3x(4x) + x2(2x2) 4. a. Find a cubic and a quadratic model for the table of values. b. Which model appears to give the better fit? c. Using the model you selected in part b, estimate the value of y when x = 12. 2x; degree 1, linear monomial 4x3 + 7x2 + 10; degree 3, cubic trinomial 2x4 + 12x2; degree 4, quartic binomial the cubic model –49.3 6-1

  12. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 (For help, go to Lessons 5-1 and 5-4.) Factor each quadratic expression. 1.x2 + 7x + 12 2.x2 + 8x – 20 3.x2 – 14x + 24 Find each product. 4.x(x + 4) 5. (x + 1)26. (x – 3)2(x + 2) 6-2

  13. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Solutions 1. Factors of 12 with a sum of 7: 4 and 3 x2 + 7x + 12 = (x + 4)(x + 3) 2. Factors of –20 with a sum of 8: 10 and –2 x2 + 8x – 20 = (x + 10)(x – 2) 3. Factors of 24 with a sum of –14: –12 and –2 x2 – 14x + 24 = (x – 12)(x – 2) 4.x(x + 4) = x(x) + x(4) = x2 + 4x 5. (x + 1)2 = x2 + 2(1)x + 12 = x2 + 2x + 1 6. (x – 3)2(x + 2) = (x2 – 2(3)x + 32)(x + 2) = (x2 – 6x + 9)(x + 2) = (x2 – 6x + 9)(x) + (x2 – 6x + 9)(2) = (x3 – 6x2 + 9x) + (2x2 – 12x + 18) = x3 + (–6 + 2)x2 + (9 – 12)x + 18 = x3 – 4x2 – 3x + 18 6-2

  14. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write (x – 1)(x + 3)(x + 4) as a polynomial in standard form. (x – 1)(x + 3)(x + 4) = (x – 1)(x2 + 4x + 3x + 12) Multiply (x + 3) and (x + 4). = (x – 1)(x2 + 7x + 12) Simplify. = x(x2 + 7x + 12) – 1(x2 + 7x + 12) Distributive Property = x3 + 7x2 + 12x – x2 – 7x – 12 Multiply. = x3 + 6x2 + 5x – 12 Simplify. The expression (x – 1)(x + 3)(x + 4) is the factored form of x3 + 6x2 + 5x – 12. 6-2

  15. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write 3x3 – 18x2 + 24x in factored form. 3x3 – 18x2 + 24x = 3x(x2 – 6x + 8) Factor out the GCF, 3x. = 3x(x – 4)(x – 2) Factor x2 – 6x + 8. Check: 3x(x – 4)(x – 2) = 3x(x2 – 6x + 8) Multiply (x – 4)(x – 2). = 3x3 – 18x2 + 24xDistributive Property 6-2

  16. Relate: Volume = depth • length • width Define: Let x = depth. Then x + 10 = length, and 50 – (depth + length) = width. Write: V(x) = x ( x + 10 )( 50 – (x + x + 10) ) = x (x + 10)(40 – 2x) The x-intercepts of the function are x = 0, x = –10, x = 20. These values of x produce a volume of zero. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Another airline has different carry-on luggage regulations. The sum of the length, width, and depth may not exceed 50 in. a. Assume that the sum of the length, width, and depth is 50 in. and the length is 10 in. greater than the depth. Graph the function relating the volume v to depth x. Find the x-intercepts. What do they represent? Graph the function for volume. 6-2

  17. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 (continued) b. Describe a realistic domain for V(x). The function has values over the set of all real numbers x. Since x represents the depth of the luggage, x > 0. Since the volume must be positive, x < 20. A realistic domain is 0 < x < 20. c. What is the maximum possible volume of the box? What are the corresponding dimensions of the box? Look for the greatest value of y that occurs within the domain 0 < x < 20. Use the Maximum feature of a graphing calculator to find the maximum volume. A volume of approximately 4225 in.3 occurs for a depth of about 12.2 in. Then the length is about 22.2 in. and the width is about 15.6 in. 6-2

  18. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Find the zeros of y = (x + 1)(x – 1)(x + 3). Then graph the function using a graphing calculator. Using the Zero Product Property, find a zero for each linear factor. x + 1 = 0    or     x – 1 = 0    or    x + 3 = 0 x = –1 x = 1 x = –3 The zeros of the function are –1, 1, –3. Now sketch and label the function. 6-2

  19. 2 –3 0Zeros ƒ(x) = (x – 2)(x + 3)(x) Write a linear factor for each zero. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Write a polynomial in standard form with zeros at 2, –3, and 0. = (x – 2)(x2 + 3x) Multiply (x + 3)(x). = x(x2 + 3x) – 2(x2 + 3x) Distributive Property = x3 + 3x2 – 2x2 – 6xMultiply. = x3 + x2 – 6xSimplify. The function ƒ(x) = x3 + x2 – 6x has zeros at 2, –3, and 0. 6-2

  20. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 Find any multiple zeros of ƒ(x) = x5 – 6x4 + 9x3 and state the multiplicity. ƒ(x) = x5 – 6x4 + 9x3 ƒ(x) = x3(x2 – 6x + 9) Factor out the GCF, x3. ƒ(x) = x3(x – 3)(x – 3) Factor x2 – 6x + 9. Since you can rewrite x3 as (x – 0)(x – 0)(x – 0), or (x – 0)3, the number 0 is a multiple zero of the function, with multiplicity 3. Since you can rewrite (x – 3)(x – 3) as (x – 3)2, the number 3 is a multiple zero of the function with multiplicity 2. 6-2

  21. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 pages 311–313  Exercises 1.x2 + x – 6 2.x3 + 12x2 + 47x + 60 3.x3 – 7x2 + 15x – 9 4.x3 + 4x2 + 4x 5.x3 + 10x2 + 25x 6.x3 – x 7.x(x – 6)(x + 6) 8. 3x(3x – 1)(x + 1) 9. 5x(2x2 – 2x + 3) 10.x(x + 5)(x + 2) 11.x(x + 4)2 12.x(x – 9)(x + 2) 13. 24.2, –1.4, 0, –5, 1 14. 5.0, –16.9, 2, 6, 8 15.a.h = x, = 16 – 2x, w = 12 – 2x b.V = x(16 – 2x)(12 – 2x) c. 194 in.3, 2.26 in. 16. 1, –2  17. 2, –9 18. 0, –5, 8 19. –1, 2, 3  6-2

  22. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 30. 0, 1 (mult. 3) 31. –1, 0, 32. –1, 0, 1 33. 4 (mult. 2) 34. 1, 2 (mult. 2) 35. – , 1 (mult. 2) 36. –1 (mult. 2), 1, 2 37. 2 x3 blocks, 15 x2 blocks, 31 x blocks, 12 unit blocks 38.a.V = 2x3 + 15x2 + 31x + 12; 2x3 + 7x2 + 7x + 2 b.V = 8x2 + 24x + 10 39.V = 12x3 – 27x 20. –1, 1, 2  21.y = x3 – 18x2 + 107x – 210 22.y = x3 + x2 – 2x 23. y = x3 + 9x2 + 15x – 25 24.y = x3 – 9x2 + 27x – 27 25. y = x3 + 2x2 – x – 2 26. y = x3 + 6x2 + 11x + 6 27.y = x3 – 2x2 28.y = x3 – x2 – 2x 29. –3 (mult. 3) 1 2 3 2 7 2 6-2

  23. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 46. y = xx – x + 47. 2.5, 5.1; , 4, 6 48. 0.9, –6.9, –1.4; 0, –3, –1, 1 49. 2.98, –6.17; 1.5 50. none, –1; –2, 0 51–53. Answers may vary. Samples are given. 51.y = x3 – 3x2 – 10x 52.y = x3 – 21x2 + 147x – 343 53.y = x4 – 4x3 – 7x2 + 22x + 24 54. –4, 5 (mult. 3) 55. 0 (mult. 2), –1 (mult. 2) 1 2 1 2 1 2 40.a.h = x + 3; w = x b. 0 < –3 < 2; where the volume is zero c. 0 < x < 2 d. 4.06 ft3 41.y = –2x3 + 9x2 – x – 12 42.y = 5x4 – 23x3 – 250x2 + 1164x + 504 43.y = 3x(x – 8)(x – 1) 44.y = –2x(x + 5)(x – 4) 45.y = x2(x + 4)(x – 1) 3 2 6-2

  24. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 56. 0, 6, –6 57. Answers may vary. Sample: Write the polynomial in standard form. The constant term is the value of the y-intercept. 58. 1 ft 59. Answers may vary. Sample: y = x4 – x2, and zeros are 0, ±1. 60. Answers may vary. Sample: The linear factors can be determined by examining the x-intercepts of the graph. 61.x + 2a 62.a.A = –x3 + 2x2 + 4x b. 6 square units 63. Answers may vary. Sample: y = (x – 1)(x + 1)(x – i )(x + i ); y = x4 – 1 64.a. Answers may vary. Sample: translation to the right 4 units b. No; the second graph is not the result of a horizontal translation. c. Answers may vary. Sample: rotation of 180° about the origin 65. C 66. H 67. B 68.[2] ƒ(x) = x2(x + 9)(x – 1) [1] incomplete factoring of x2 + 8x – 9 or other minor errors 7 8 6-2

  25. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 69.[4] The student multiplies (x + 2), (x – 5), (x – 6) and –2 to get y = –2x3 + 18x2 – 16x – 120. [3] appropriate methods but with one computational error [2] appropriate methods to find basic polynomial with given roots, but no multiplication of polynomial by –2 [1] correct function, without work shown 70. –3x5 + 3x2 – 1; quintic trinomial 71. –7x4 – x3; quartic binomial 72.x3 – 2; cubic binomial 73. (x + 4)(x + 1) 74. (x – 5)(x + 3) 75. (x – 6)2 76. 8 77. –11 78. 0 6-2

  26. 400 27 x(x + 3)(x – 5); , –36 Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 1. Find the zeros of y = (x + 7)(x – 3)(x – 2). Then write the polynomial in standard form. 2. Factor x3 – 2x2 – 15x. Find the relative maximum and relative minimum. 3. Write a polynomial function in standard form with zeros at –5, –4, and 3. 4. Find any multiple zeros of ƒ(x) = x4 – 25x2 and state the multiplicity. –7, 2, 3; x3 + 2x2 – 29x + 42 Answers may vary. Sample: ƒ(x) = x3 + 6x2 – 7x – 60 The number 0 is a zero with multiplicity 2. 6-2

  27. Dividing Polynomials ALGEBRA 2 LESSON 6-3 (For help, go to Lessons 5-1 and 6-1.) Simplify each expression. 1. (x + 3)(x – 4) + 2 2. (2x + 1)(x – 3) 3. (x + 2)(x + 1) – 11 4. –3(2 – x)(x + 5) Write each polynomial in standard form. Then list the coefficients. 5. 5x – 2x2 + 9 + 4x36. 10 + 5x3 – 9x2 7. 3x + x2 – x + 7 – 2x28. –4x4 – 7x2 + x3 + x4 6-3

  28. Dividing Polynomials ALGEBRA 2 LESSON 6-3 Solutions 1. (x + 3)(x – 4) + 2 = x2 – 4x + 3x – 12 + 2 = x2 – x – 10 2. (2x + 1)(x – 3) = 2x2 – 6x + x – 3 = 2x2 – 5x – 3 3. (x + 2)(x + 1) – 11 = x2 + x + 2x + 2 – 11 = x2 + 3x – 9 4. –3(2 – x)(x + 5) = –3(2x + 10 – x2 – 5x) = –3(–x2 – 3x + 10) = 3x2 + 9x – 30 5. 5x – 2x2 + 9 + 4x3 = 4x3 – 2x2 + 5x + 9 coefficients: 4, –2, 5, 9 6. 10 + 5x3 – 9x2 = 5x3 – 9x2 + 10 coefficients: 5, –9, 0, 10 7. 3x + x2 – x + 7 – 2x2 = –x2 + 2x + 7 coefficients: –1, 2, 7 8. –4x4 – 7x2 + x3 + x4 = –3x4 + x3 – 7x2 coefficients: –3, 1, –7, 0, 0 6-3

  29. x2 x xDivide = x. x – 5x2 + 2x – 30 7x x x2 – 5xMultiply: x(x – 5) = x2 – 5x 7x x+ 7Divide = 7. x – 5x2 + 2x – 30 x2 – 5x 7x – 30 7x – 35 Multiply: 7(x – 5) = 7x – 35. Dividing Polynomials ALGEBRA 2 LESSON 6-3 Divide x2 + 2x – 30 by x – 5.  – 30Subtract: (x2 + 2x) – (x2 – 5x) = 7x. Bring down –30. Repeat the process of dividing, multiplying, and subtracting. 5 Subtract: (7x – 30) – (7x – 35) = 5. The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5. 6-3

  30. x2 + 5x – 15 x + 2 x3 + 7x2 – 5x – 6 x3 + 2x2 5x2 – 5x 5x2 + 10x –15x – 6 –15x – 30 24 x + 8 x + 2 x2 + 10x + 16 x2 + 2x 8x + 16 8x + 16 0 Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6. / Dividing Polynomials ALGEBRA 2 LESSON 6-3 Determine whether x + 2 is a factor of each polynomial. a.x2 + 10x + 16 b.x3 + 7x2 – 5x – 6 Since the remainder is zero, x + 2 is a factor of x2 + 10x + 16. 6-3

  31. Write x – 3 5x3 – 6x2 + 4x – 1 as 3 5 –6 4 –1 Bring down the 5. 3 5 –6 4 –1 This begins the quotient. 5 Dividing Polynomials ALGEBRA 2 LESSON 6-3 Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3. Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form. Step 2: Bring down the coefficient. 6-3

  32. Multiply 3 by 5. Write 35 –6 4 –1 the result under –6. x15 5 9 Add –6 and 15. 3 5 –6 4 –1 15 27 93 5 9 31 92 5x2 + 9x + 31  Remainder Dividing Polynomials ALGEBRA 2 LESSON 6-3 (continued) Step 3:  Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add. Step 4:  Repeat the steps of multiplying and adding until the remainder is found. The quotient is 5x2 + 9x + 31, R 92. 6-3

  33. 5 1 –6 3 10 Divide. 5 –5 –10 1 –1 –2 0 x2 – x – 2  Remainder Dividing Polynomials ALGEBRA 2 LESSON 6-3 The volume in cubic feet of a shipping carton is V(x) = x3 – 6x2 + 3x + 10. The height is x – 5 feet. a. Find linear expressions for the other dimensions. Assume that the length is greater than the width. x2 – x – 2 = (x – 2)(x + 1) Factor the quotient. The length and the width are x + 1 and x – 2, respectively. b. If the width of the carton is 4 feet, what are the other two dimensions? x – 2 = 4  Substitute 4 into the expression for width. Find x. x = 6 Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft. 6-3

  34. 3 1 –2 0 1 –9 3 3 9 30 1 1 3 10 21 Dividing Polynomials ALGEBRA 2 LESSON 6-3 Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9. By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3. The remainder is 21, so P(3) = 21. 6-3

  35. Dividing Polynomials ALGEBRA 2 LESSON 6-3 pages 318–320  Exercises 1.x – 8 2. 3x – 5 3.x2 + 4x + 3, R 5 4. 2x2 + 5x + 2 5. 3x2 – 7x + 2 6. 9x – 12, R –32 7.x – 10, R 40 8.x2 + 4x + 3 9. no 10. yes 11. yes 12. no 13.x2 + 4x + 3 14.x2 – 2x + 2 15. x2 – 11x + 37, R –128 16. x2 + 2x + 5 17. x2 – x – 6 18. –2x2 + 9x – 19, R 40 19.x + 1, R 4 20. 3x2 + 8x – 3 21.x2 – 3x + 9 22. 6x – 2, R –4 23.y = (x + 1)(x + 3)(x – 2) 6-3

  36. Dividing Polynomials ALGEBRA 2 LESSON 6-3 24.y = (x + 3)(x – 4)(x – 3) 25. = x + 3 and h = x 26. 18 27. 0 28. 0 29. 12 30. 168 31. 10 32. 51 33. 0 34. P(a) = 0; x – a is a factor of P(x). 35.x – 1 is not a factor of x3 – x2 – 2x because it does not divide into x3 – x2 – 2x evenly. 36. Answers may vary. Sample: (x2 + x – 4) ÷ (x – 2) 37.x2 + 4x + 5 38. x3 – 3x2 + 12x – 35, R 109 39. x4 – x3 + x2 – x + 1 40. x + 4 41.x3 – x2 + 1 42. no 43. yes 44. yes 6-3

  37. Dividing Polynomials ALGEBRA 2 LESSON 6-3 45. no 46. no 47. yes 48. yes 49. yes 50. no 51. no 52.x3 – x2 + 1 53.x3 – 2x2 – 2x + 4, R –35 54.x3 – 2x2 – x + 6 55.x3 – 4x2 + x 56.a.x + 1 b.x2 + x + 1 c.x3 + x2 + x + 1 d. (x – 1)(x4 + x3 + x2 + x + 1) 57. a.x2 – x + 1 b.x4 – x3 + x2 – x + 1 c.x6 – x5 + x4 – x3 + x2 – x + 1 d. (x + 1)(x8 – x7 + x6 – x5 + x4 – x3 + x2 – x + 1) 58. By dividing it by a polynomial of degree 1, you are reducing the degree-n polynomial by one, to n – 1. The remainder will be constant because it is not divisible by the variable. 59.x + 2i 6-3

  38. Dividing Polynomials ALGEBRA 2 LESSON 6-3 60. Yes; the graph could rise to the right and fall to the left or it could fall to the right and rise to the left. 61. D 62. I 63. B 64.[2]x2 – 6x – 7 x – 1 x3 – 7x2 – x + 7 x3 – x2 –6x2 – x –6x2 + 6x –7x + 7 – 7x + 7 0 x2 – 6x – 7 = 0 (x + 1)(x – 7) = 0 x = 1, –1, 7 [1] incorrect illustration of division or incorrect list of zeros 65.y = x2 + 2x – 15 66.y = x3 –9x2 + 8x 67. y = x3 – 6x2 + 3x + 10 68.y = x4 – 4x3 + 6x2 – 4x + 1 69. 24 70. 5 71. 23 – 11i 6-3

  39. Dividing Polynomials ALGEBRA 2 LESSON 6-3 72. –0.5 0 –1.5 0.5 1 1.5 0.5 0 0.5 73. none exists 74. 0 2 1 1 –4 –2 0.5 –3.5 –1.5 6-3

  40. Dividing Polynomials ALGEBRA 2 LESSON 6-3 1a. Divide using long division. (9x3 – 48x2 + 13x + 3) ÷ (x – 5) 1b. Is x – 5 a factor of 9x3 – 48x2 + 13x + 3? 2. Divide using synthetic division. (6x3 – 4x2 + 14x – 8) ÷ (x + 2) 3. Use synthetic division and the given factor x – 4 to completely factor x3 – 37x + 84. 4. Use synthetic division and the Remainder Theorem to find P(–2) when P(x) = x4 – 2x3 + 4x2 + x + 1. 9x2 – 3x – 2, R –7 no 6x2 – 16x + 46, R –100 (x + 7)(x – 3)(x – 4) 47 6-3

  41. Graph each system. Find any points of intersection. 1. 2. 3. Factor each expression. 4.x2 – 2x – 15 5.x2 – 9x + 14 6.x2 + 6x + 5 y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 2y = – x + 8 x + 2y = –6 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (For help, go to Lessons 6-2 and 6-3.) 6-4

  42. y = 3x + 1 y = –2x + 6 –2x + 3y = 0 x + 3y = 3 2y = – x + 8 x + 2y = –6 2 3 y = x y = – x + 1 1 3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Solutions 1.    2. intersection: (1, 4) 3.     Rewrite equations: Rewrite equations: intersection: (1, ) no points of intersection 4. Factors of –15 with a sum of –2: –5 and 3 x2 – 2x – 15 = (x – 5)(x + 3) 5. Factors of 14 with a sum of –9: –7 and –2 x2 – 9x + 14 = (x – 7)(x – 2) 6. Factors of 5 with a sum of 6: 5 and 1 x2 + 6x + 5 = (x + 5)(x + 1) 1 2 y = – x + 4 y = – x – 3 2 3 1 2 6-4

  43. Step 1:  Graph y1 = x3 – 19x and y2 = –2x2 + 20 on a graphing calculator. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 Graph and solve x3 – 19x = –2x2 + 20. Step 2:  Use the Intersect feature to find the x values at the points of intersection. The solutions are –5, –1, and 4. 6-4

  44. Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 (continued) Check: Show that each solution makes the original equation a true statement. x3 – 19x = –2x2 + 20 x3 – 19x = –2x2 + 20 (–5)3 – 19(–5) –2(–5)2 + 20  (–1)3 – 19(–1) –2(–1)2 + 20 –125 + 95 –50 + 20 –1 + 19 –2 + 20 –30 = –30 18 = 18 x3 – 19x = –2x2 + 20 (4)3 – 19(4) –2(4)2 + 20 64 – 76 –32 + 20 –12 = –12 6-4

  45. 15.9 ft3 • = 27475.2 in.3Convert the volume to cubic inches. Graph y1 = 27475.2 and y2 = (x + 4)x(x – 3).    Use the Intersect option of the calculator. When y = 27475.2, x 30. So x – 3 27 and x + 4 34. 123 in.3 ft3 Solving Polynomial Equations ALGEBRA 2 LESSON 6-4 The dimensions in inches of the cubicle area inside a doghouse can be expressed as width x, length x + 4, and height x – 3. The volume is 15.9 ft3. Find the dimensions of the doghouse. V = l • w • hWrite the formula for volume. 27475.2 = (x + 4)x(x – 3) Substitute. The dimensions of the doghouse are about 30 in. by 27 in. by 34 in. 6-4

  46. 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

  47. 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

  48. –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

  49. 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

  50. 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