Study Guide and Review - Chapter 2

Transcription

1 Divide using long division. 31. (x 3 + 8x 2 5) (x 2) So, (x 3 + 8x 2 5) (x 2) = x x (2x 5 + 5x 4 5x 3 + x 2 18x + 10) (2x 1) So, (2x 5 + 5x 4 5x 3 + x 2 18x + 10) (2x 1) = x 4 + 3x 3 x esolutions Manual - Powered by Cognero Page 1

2 Divide using synthetic division. 35. (x 4 x 3 + 7x 2 9x 18) (x 2) Because x 2, c = 2. Set up the synthetic division as follows. Then follow the synthetic division procedure. The quotient is x 3 + x 2 + 9x + 9. 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) = x 3 + 3x 2 8x 24; (x + 3) Use synthetic division to test the factor (x + 3). 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) = (x + 3)(x 2 8). 39. f (x) = x 4 2x 3 3x 2 + 4x + 4; (x + 1), (x 2) Use synthetic division to test each factor, (x + 1) and (x 2). Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x 2), with the depressed polynomial x 3 3x 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 f (x) as f (x) = (x + 1)(x 2)(x 2 x 2). Factoring the quadratic expression yields f (x) = (x 2) 2 (x + 1) 2. esolutions Manual - Powered by Cognero Page 2

3 List all possible rational zeros of each function. Then determine which, if any, are zeros. 41. f (x) = x 3 14x 15 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) = (x + 3)(x 2 3x 5). Because the factor (x 2 3x 5) yields no rational zeros, the rational zero of f is f (x) = 3x 4 14x 3 2x x + 10 The leading coefficient is 3 and the constant term is 10. The possible rational zeros are 1, 2, 5, 10,. By using synthetic division on the depressed polynomial, it can be determined that is a rational zero. Because (x 2) and are factors of f (x), we can use the final quotient to write a factored form of f (x) as Since (3x 2 9x 15) yields no rational zeros, the rational zeros of f are esolutions Manual - Powered by Cognero Page 3

4 Solve each equation x 3 23x x 8 = 0 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)(6x 2 11x + 4) = 0. Factoring the quadratic expression yields (x 2)(3x 4)(2x 1) = 0.Thus, the solutions are 2,, and x 4 11x x = 4x The equation can be written as 2x 4 11x 3 + 4x x 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 on the depressed polynomial, it can be determined that 4 is a rational zero. Because (x 2) and (x 4) are factors of the equation, we can use the final quotient to write a factored form as (x 2)(x 4)(2x 2 + 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. esolutions Manual - Powered by Cognero Page 4

5 Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. f (x) = x 4 4x 3 + 7x 2 16x + 12; 2i 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 from the last division, write f (x) = (x + 2i)(x 2i)(x 2 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 5