Explanation Formula used: 1) Theorem: Rational Zeros: Let f ( x ) = a 0 + a 1 x + ... + a n − 1 x n − 1 + a n x n be a polynomial of positive degree n with coefficients that are all integers, and let p / q be a rational number that has been written in its lowest terms. If p / q is a zero of f ( x ) , then p divides a 0 and q divides a n . 2) Quadratic formula: The zeros of a polynomial f ( x ) = a x 2 + b x + c with real coefficients and a ≠ 0 are given by, r 1 = − b + b 2 − 4 a c 2 a and r 2 = − b − b 2 − 4 a c 2 a . Explanation: Let, f ( x ) = 9 x 3 + 27 x 2 + 8 x − 20 . To find all the zeros of f ( x ) First, finding the rational zeros of f ( x ) . By the theorem, any rational zero p / q of f ( x ) that is a lowest term must have a numerator p that divides the constant term and a denominator q that divides the leading coefficient. That is, p | ( − 20 ) ⇒ p ∈ { ± 1 , ± 2 , ± 4 , ± 5 , ± 10 , ± 20 } , and q | 9 ⇒ q ∈ { ± 1 , ± 3 , ± 9 } . So, p q ∈ { ± 1 9 , ± 2 9 , ± 1 3 , ± 2 3 , ± 4 9 , ± 5 9 , ± 4 3 , ± 5 3 , ± 10 9 , ± 20 9 , ± 10 3 , ± 20 3 , ± 1 , ± 2 , ± 4 , ± 5 , ± 10 , ± 20 } . Now, testing the possibilities of x = p / q , For, x = 1 9 ⇒ f ( 1 9 ) = 9 ( 1 9 ) 3 + 27 ( 1 9 ) 2 + 8 ( 1 9 ) − 20 = 9 ( 1 729 ) + 27 ( 1 81 ) + 8 ( 1 9 ) − 20 = − 1520 81 ∴ f ( 1 9 ) = − 1520 81 ≠ 0 . For, x = − 1 9 ⇒ f ( − 1 9 ) = 9 ( − 1 9 ) 3 + 27 ( − 1 9 ) 2 + 8 ( − 1 9 ) − 20 = 9 ( − 1 729 ) + 27 ( 1 81 ) − 8 ( 1 9 ) − 20 = − 1666 81 ∴ f ( − 1 9 ) = − 1666 81 ≠ 0 . For, x = 2 9 ⇒ f ( 2 9 ) = 9 ( 2 9 ) 3 + 27 ( 2 9 ) 2 + 8 ( 2 9 ) − 20 = 9 ( 8 729 ) + 27 ( 4 81 ) + 8 ( 2 9 ) − 20 = − 1360 81 ∴ f ( 2 9 ) = − 1360 81 ≠ 0

8th Edition

ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Chapter 8.4, Problem 12E

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All zeros of the polynomial 9x3+27x2+8x−20.

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Chapter 8.4, Problem 11E

Chapter 8.4, Problem 13E

Chapter 8 Solutions

Elements Of Modern Algebra

Ch. 8.1 - True or False Label each of the following...Ch. 8.1 - Prob. 2TFE Ch. 8.1 - Prob. 3TFE Ch. 8.1 - Prob. 4TFE Ch. 8.1 - Prob. 5TFE Ch. 8.1 - Prob. 6TFE Ch. 8.1 - Prob. 7TFE Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E

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All zeros of the polynomial 9 x 3 + 27 x 2 + 8 x − 20 .

Solution Summary: The author explains the rational zeros of the polynomial f(x) and the quadratic formula for finding them.

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All zeros of the polynomial 9x3+27x2+8x−20.

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