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Polynomials and Linear Factors

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 x 3 – 2 x 2 – 15 x . Find the relative maximum and relative minimum.

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Polynomials and Linear Factors

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  1. 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

  2. Assignment 52 • Page 311 16-36 even, 52-55

  3. 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

  4. 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

  5. Polynomials and Linear Factors ALGEBRA 2 LESSON 6-2 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) 6-2

  6. Dividing Polynomials • Recall that when a numerical division has a remainder of zero, as shown below, the divisor and quotient are both factors of the dividend. • Both 4 and 6 are factors of 24

  7. Dividing Polynomials • The same is true for polynomial division. If you divide a polynomial by one of its factors, then you get another factor. When a polynomial division leaves no remainder, as shown below with monomials, you have factored the polynomial.

  8. Long Division Review

  9. 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

  10. 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

  11. Assignment 54 • Page 318 1-12, 37-40

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