24 Zeros of Polynomial Functions


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1 Write a polynomial function of least degree with real coefficients in standard form that has the given zeros , 4, 3, 5 Using the Linear Factorization Theorem and the zeros 2, 4, 3, and 5, write f (x) as follows. f(x) = a[x ( 2)][x ( 4)][x ( 3)][x (5)] Therefore, a function of least degree that has 2, 4, 3, and 5 as zeros is f (x) = x 4 + 4x 3 19x 2 106x 120 or any nonzero multiple of f (x) , 8, 6 i Because 6 i is a zero and the polynomial is to have real coefficients, you know that 6 + i must also be a zero. Using the Linear Factorization Theorem and the zeros 1, 8, 6 i, and 6 + i, write f (x) as follows. f(x) = a[x ( 1)][x (8)][x (6 i)][x (6 + i)] Therefore, a function of least degree that has 1, 8, 6 i, and 6 + i as zeros is f (x) = x 4 19x x 2 163x 296 or any nonzero multiple of f (x) , 2, 4, 4 + Using the Linear Factorization Theorem and the zeros 5, 2, 4, and 4 +, write f (x) as follows. f(x) = a[x ( 5)][x (2)][x (4 )][x (4 + )] Therefore, a function of least degree that has 5, 2, 4, and 4 + as zeros is f (x) = x 4 5x 3 21x x 130 or any nonzero multiple of f (x). esolutions Manual  Powered by Cognero Page 1
2 39.,, 3 4i Because 3 4i is a zero and the polynomial is to have real coefficients, you know that 3 + 4i must also be a zero. Using the Linear Factorization Theorem and the zeros,, 3 4i, and 3 + 4i, write f (x) as follows. f(x) = a[x ( )][x ( )][x (3 4i)][x (3 + 4i)] Therefore, a function of least degree that has,, 3 4i, and 3 + 4i as zeros is f (x) = x 4 6x x x 150 or any nonzero multiple of f (x) , 6 +, 8 3i Because 8 3i is a zero and the polynomial is to have real coefficients, you know that 8 + 3i must also be a zero. Using the Linear Factorization Theorem and the zeros 6, 6 +, 8 3i, and 8 + 3i, write f (x) as follows. f(x) = a[x (6 )][x (6 + )][x (8 3i)][x (8 + 3i)] Therefore, a function of least degree that has 6, 6 +, 8 3i, and 8 + 3i as zeros is f (x) = x 4 28x x x or any nonzero multiple of f (x). esolutions Manual  Powered by Cognero Page 2
3 Write each function as (a) the product of linear and irreducible quadratic factors and (b) the product of linear factors. Then (c) list all of its zeros. 43. g(x) = x 4 3x 3 12x a. g(x) has possible rational zeros of ±1, ±2, ±4, ±8. By using synthetic division, it can be determined that x = 2 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that x = 1 is a rational zero. The remaining quadratic factor (x 2 6x + 4) yields no rational zeros. Use the quadratic formula to find the zeros. So, g(x) written as a product of linear and irreducible quadratic factors is g(x) = (x + 2)(x + 1)(x 3 + ). b. g(x) written as a product of linear factors is g(x) = (x + 2)(x + 1)(x 3 + )(x 3 ). c. The zeros are 2, 1, 3, 3 +. )(x 3 esolutions Manual  Powered by Cognero Page 3
4 45. f (x) = 4x 4 35x x 2 295x a. f (x) has possible rational zeros of ±1, ±2, ±3, ±4, ±12, ±13, ±26, ±39, ±52, ±78, ±156, By using synthetic division, it can be determined that x = 4 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that x = is a rational zero. The remaining quadratic factor (4x 2 16x + 52) can be written 4(x 2 4x + 13) and yields no real zeros and is therefore, irreducible over the reals. So, f (x) written as a product of linear and irreducible quadratic factors is or (x 2 4x + 13)(4x 3)(x 4) b. Use the quadratic formula to find the zeros of x 2 4x x 2 16x + 52 can be written as [x (2 + 3i)][x (2 3i)]. Thus, f (x) written as a product of linear factors is f (x) = (4x 3)(x 4)(x 2 + 3i)(x 2 3i). c. The zeros are, 4, 2 3i, and 2 + 3i. esolutions Manual  Powered by Cognero Page 4
5 47. h(x) = x 4 2x 3 17x 2 + 4x + 30 a. h(x) has possible rational zeros of ±1, ±3, ±5, ±6, ±10, and ±30. By using synthetic division, it can be determined that x = 3 is a rational zero. By using synthetic division on the depressed polynomial, it can be determined that x = 5 is a rational zero. The remaining quadratic factor (x 2 2) yields no rational zeros and can be written as (x + )(x ). So, h(x) written as a product of linear and irreducible quadratic factors is h(x) = (x + 3)(x 5)(x + )(x ). b. h(x) written as a product of linear factors is h(x) = (x + 3)(x 5)(x + )(x ). c. The zeros are 3, 5,, and. Use the given zero to find all complex zeros of each function. Then write the linear factorization of the function. 49. h(x) = 2x 5 + x 4 7x x 2 225x + 108; 3i Use synthetic substitution to verify that 3i is a zero of h(x). Because x = 3i is a zero of h, x = 3i is also a zero of h. Divide the depressed polynomial by 3i. Using these two zeros and the depressed polynomial from the last division, write h(x) = (x + 3i)(x 3i)(2x 3 + x 2 25x + 12). 2x 3 + x 2 25x + 12 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±12,. By using synthetic division, it can be determined that x = 3 is a rational zero. The remaining depressed polynomial 2x 2 +7x 4 can be written as (x + 4)(2x 1). The zeros of the depressed polynomial are 4 and. Therefore, the zeros of h are 3, 4,, 3i, and 3i. The linear factorization of h is h(x) = (x 3)(x + 4)(2x 1)(x + 3i)(x 3i). esolutions Manual  Powered by Cognero Page 5
6 51. g(x) = x 5 2x 4 13x x x 60; 3 i Use synthetic substitution to verify that 3 i is a zero of g(x). Because x = 3 i is a zero of g, x = 3 + i is also a zero of g. Divide the depressed polynomial by 3 + i. Using these two zeros and the depressed polynomial from the last division, write h(x) = [x (3 i)][x (3 + i)](x 3 + 4x 2 + x 6). x 3 + 4x 2 + x 6 has possible rational zeros of ±1, ±2, ±3, and ±6. By using synthetic division, it can be determined that x = 1 is a rational zero. The remaining depressed polynomial x 2 + 5x + 6 can be written as (x + 3)(x + 2). The zeros of the depressed polynomial are 3 and 2. Therefore, the zeros of g are 3, 2, 1, 3 + i, and 3 i. The linear factorization of g is g (x) = (x + 3)(x + 2)(x 1)(x 3 + i)(x 3 i). 53. f (x) = x 5 3x 4 4x x 2 32x + 96; 2i Use synthetic substitution to verify that 2i is a zero of f (x). Because x = 2i is a zero of f, x = 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 3 3x 2 8x + 24). x 3 3x 2 8x + 24 has possible rational zeros of ±1, ±2, ±3, ±4, ±6, ±8, ±12, and ±24. By using synthetic division, it can be determined that x = 3 is a rational zero. The remaining depressed polynomial x 2 8 can be written as (x )(x + ) or (x 2 )(x + 2 ). The zeros of the depressed polynomial are 2 and 2. Therefore, the zeros of f are 3, 2, 2, 2i, and 2i. The linear factorization of f is f (x) = (x 3)(x + 2 )(x 2 )(x + 2i)(x 2i). esolutions Manual  Powered by Cognero Page 6
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