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 ) = 6 x 3 + 11 x 2 + x − 4 . 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 | ( − 4 ) ⇒ p ∈ { ± 1 , ± 2 , ± 4 } , and q | 6 ⇒ q ∈ { ± 1 , ± 2 , ± 3 , ± 6 } . So, p q ∈ { ± 1 2 , ± 1 3 , ± 1 6 , ± 1 , ± 2 , ± 2 3 , ± 4 , , ± 4 3 , }

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ISBN: 9781285463230

Author: Gilbert, Linda, Jimmie

Publisher: Cengage Learning,

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Try substituting 1. You will get 6 and not 0. Try other real values also. What About -2? Let us check.

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Chapter 8.4, Problem 11E

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All zeros of the polynomial 6x3+11x2+x−4.

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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 6 x 3 + 11 x 2 + x − 4 .

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 6x3+11x2+x−4.

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