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5 4 3 2 So, (2x + 5x − 5x + x − 18x + 10) ÷ (2x – 1) = Study Guide and Review - Chapter 2 Divide using long division. 3 4 3 2 x + 3x − x − 9 + . Divide using synthetic division. 2 35. (x4 – x3 + 7x2 – 9x – 18) ÷ (x – 2) 31. (x + 8x – 5) ÷ (x – 2) SOLUTION: SOLUTION: Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure. 3 2 The quotient is x + x + 9x + 9. 3 2 2 So, (x + 8x – 5) ÷ (x – 2) = x + 10x + 20 + . Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x). 37. f (x) = x3 + 3x2 – 8x – 24; (x + 3) SOLUTION: 33. (2x5 + 5x4 − 5x3 + x2 − 18x + 10) ÷ (2x – 1) Use synthetic division to test the factor (x + 3). SOLUTION: Because the remainder when the polynomial is divided by (x + 3) is 0, (x + 3) is a factor of f (x). Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) 2 = (x + 3)(x – 8). 39. f (x) = x4 – 2x3 – 3x2 + 4x + 4; (x + 1), (x – 2) SOLUTION: Use synthetic division to test each factor, (x + 1) and (x − 2). 5 4 3 2 So, (2x + 5x − 5x + x − 18x + 10) ÷ (2x – 1) = 4 3 2 x + 3x − x − 9 + . Divide using synthetic division. Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x 3 2 − 2), with the depressed polynomial x − 3x + 4. 35. (x4 – x3 + 7x2 – 9x – 18) ÷ (x – 2) SOLUTION: Because x − 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure. eSolutions Manual - Powered by Cognero Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x). Because (x + 1) and (x − 2) are factors of f (x), we Page 1 can use the final quotient to write a factored form of 2 3 2 The quotient is x + x + 9x + 9. f(x) as f (x) = (x + 1)(x − 2)(x − x − 2). Factoring 2 divided by (x + 3) is 0, (x + 3) is a factor of f (x). Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) 2 Study = (xGuide + 3)(x –and 8). Review - Chapter 2 39. f (x) = x4 – 2x3 – 3x2 + 4x + 4; (x + 1), (x – 2) Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) 2 2 = (x + 3)(x −3x − 5). Because the factor (x − 3x − 5) yields no rational zeros, the rational zero of f is −3. 43. f (x) = 3x4 – 14x3 – 2x2 + 31x + 10 SOLUTION: SOLUTION: Use synthetic division to test each factor, (x + 1) and (x − 2). The leading coefficient is 3 and the constant term is 10. The possible rational zeros are 1, 2, 5, 10, Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x 3 2 − 2), with the depressed polynomial x − 3x + 4. . By using synthetic division, it can be determined that 2 is a rational zero. By using synthetic division on the depressed Because the remainder when the depressed polynomial is divided by (x − 2) is 0, (x − 2) is a factor of f (x). Because (x + 1) and (x − 2) are factors of f (x), we can use the final quotient to write a factored form of polynomial, it can be determined that rational zero. is a 2 f(x) as f (x) = (x + 1)(x − 2)(x − x − 2). Factoring 2 the quadratic expression yields f (x) = (x – 2) (x + 1) 2 . Because (x − 2) and List all possible rational zeros of each function. Then determine which, if any, are zeros. 41. f (x) = x3 – 14x − 15 SOLUTION: are factors of f (x), we can use the final quotient to write a factored form of Since f(x) as 2 Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term −15. Therefore, the possible rational zeros of f are 1, 3, 5, and 15. By using synthetic division, it can be determined that –3 is a rational zero. Because (x + 3) is a factor of f (x), we can use the final quotient to write a factored form of f (x) as f (x) 2 2 = (x + 3)(x −3x − 5). Because the factor (x − 3x − 5) yields no rational zeros, the rational zero of f is −3. (3x − 9x − 15) yields no rational zeros, the rational zeros of f are Solve each equation. 45. 6x3 – 23x2 + 26x – 8 = 0 SOLUTION: Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is −8. The possible rational zeros are 1, 2, 4, 8, . By using synthetic division, it can be determined that 2 is a rational zero. 43. f (x) = 3x4 – 14x3 – 2x2 + 31x + 10 SOLUTION: eSolutions - Powered by Cognero TheManual leading coefficient is 3 and the constant term is 10. The possible rational zeros are 1, 2, 5, 10, . Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form asPage (x −2 2 2)(6x − 11x + 4) = 0. Factoring the quadratic expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus, Since f(x) as 2 (3x − 9x − 15) yields no rational zeros, the rational the solutions are 2, Study Guide zeros of f areand Review - Chapter 2 , and . 47. 2x4 – 11x3 + 44x = –4x2 + 48 Solve each equation. 45. 6x3 – 23x2 + 26x – 8 = 0 SOLUTION: SOLUTION: 4 Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 6 and the constant term is −8. The possible rational zeros are 1, 2, 4, 8, . Because (x − 2) is a factor of the equation, we can use the final quotient to write a factored form as (x − 2 2)(6x − 11x + 4) = 0. Factoring the quadratic expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus, , and . 3 2 The equation can be written as 2x – 11x + 4x + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros are ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± By using synthetic division, it can be determined that 2 is a rational zero. the solutions are 2, 2)(6x − 11x + 4) = 0. Factoring the quadratic expression yields (x − 2)(3x − 4)(2x − 1) = 0.Thus, , and ± . By using synthetic division, it can be determined that 2 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that 4 is a rational zero. 47. 2x4 – 11x3 + 44x = –4x2 + 48 SOLUTION: 4 3 2 The equation can be written as 2x – 11x + 4x + 44x − 48 = 0. Apply the Rational Zeros Theorem to find possible rational zeros of the equation. The leading coefficient is 2 and the constant term is −48. The possible rational zeros are ±1, ±2, ±3, ±6, ±8, ±16, ±24, ±48, ± , and ± . By using synthetic division, it can be determined that 2 is a rational zero. Because (x − 2) and (x − 4) are factors of the equation, we can use the final quotient to write a 2 factored form as (x − 2)(x − 4)(2x + x − 6) = 0. Factoring the quadratic expression yields (x − 2)(x − 4)(2x − 3)(x + 2) = 0. Thus, the solutions are , –2, 2, and 4. Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. f (x) = x4 – 4x3 + 7x2 − 16x + 12; −2i SOLUTION: Use synthetic substitution to verify that −2i is a zero of f (x). By using synthetic division on the depressed polynomial, it can be determined that 4 is a rational zero. Because −2i is a zero of f , 2i is also a zero of f . Divide the depressed polynomial by 2i. eSolutions Manual(x - Powered by (x Cognero Because − 2) and − 4) are factors of the equation, we can use the final quotient to write a 2 factored form as (x − 2)(x − 4)(2x + x − 6) = 0. Page 3 factored form as (x − 2)(x − 4)(2x + x − 6) = 0. Factoring the quadratic expression yields (x − 2)(x − 4)(2x − 3)(x + 2) = 0. Thus, the solutions are Study Guide and Review - Chapter 2 , –2, 2, and 4. Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. f (x) = x4 – 4x3 + 7x2 − 16x + 12; −2i SOLUTION: Use synthetic substitution to verify that −2i is a zero of f (x). Because −2i is a zero of f , 2i is also a zero of f . Divide the depressed polynomial by 2i. Using these two zeros and the depressed polynomial 2 from the last division, write f (x) = (x + 2i)(x − 2i)(x − 4x + 3). Factoring the quadratic expression yields f (x) = (x + 2i)(x − 2i)(x − 3)(x − 1).Therefore, the zeros of f are 1, 3, 2i, and −2i. eSolutions Manual - Powered by Cognero Page 4