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 the coefficients that are all integers, and let p / q be a rational number that has been written in lowest terms. If p / q is a zero of f ( x ) , then p divides a 0 and q divides a n . 2) Eisenstein’s Irreducibility Criterion: Let f ( x ) = a 0 + a 1 x + ... + a n − 1 x n − 1 + a n x n be a polynomial of positive degree with integral coefficients. If there exists a prime integer p , such that p | a i for i = 0 , 1 , . . . , n − 1 , but p | a n and p 2 | a 0 , then f ( x ) is irreducible over the field of rational numbers. Explanation: Let, f ( x ) = 2 x 4 + 5 x 3 − 7 x 2 − 10 x + 6 . To find the factor of f ( x ) , that is to find 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 the lowest term must have a numerator p that divides the constant term and a denominator q that divides the leading coefficient. That is, p | 6 ⇒ p ∈ { ± 1 , ± 2 , ± 3 , ± 6 } , and q | 2 ⇒ q ∈ { ± 1 , ± 2 } . So, p q ∈ { ± 1 2 , ± 1 , ± 2 , ± 3 2 , ± 3 , ± 6 } . Now, testing the possibilities of x = p / q , For, x = 1 2 ⇒ f ( 1 2 ) = 2 ( 1 2 ) 4 + 5 ( 1 2 ) 3 − 7 ( 1 2 ) 2 − 10 ( 1 2 ) + 6 = 2 ( 1 16 ) + 5 ( 1 8 ) − 7 ( 1 4 ) − 10 ( 1 2 ) + 6 = 0 ∴ f ( 1 2 ) = 0 . For, x = − 1 2 ⇒ f ( − 1 2 ) = 2 ( − 1 2 ) 4 + 5 ( − 1 2 ) 3 − 7 ( − 1 2 ) 2 − 10 ( − 1 2 ) + 6 = 2 ( 1 16 ) − 5 ( 1 8 ) − 7 ( 1 4 ) + 10 ( 1 2 ) + 6 = 35 4 ∴ f ( − 1 2 ) = 35 4 ≠ 0 . For, x = 1 ⇒ f ( 1 ) = 2 ( 1 ) 4 + 5 ( 1 ) 3 − 7 ( 1 ) 2 − 10 ( 1 ) + 6 = 2 + 5 − 7 − 10 + 6 = − 4 ∴ f ( 1 ) = − 4 . For, x = − 1 ⇒ f ( − 1 ) = 2 ( − 1 ) 4 + 5 ( − 1 ) 3 − 7 ( − 1 ) 2 − 10 ( − 1 ) + 6 = 2 − 5 − 7 + 10 + 6 = 6 ∴ f ( − 1 ) = 6 . For, x = 2 ⇒ f ( 2 ) = 2 ( 2 ) 4 + 5 ( 2 ) 3 − 7 ( 2 ) 2 − 10 ( 2 ) + 6 = 32 + 40 − 28 − 20 + 6 = 30 ∴ f ( 2 ) = 30 . For, x = − 2 ⇒ f ( − 2 ) = 2 ( − 2 ) 4 + 5 ( − 2 ) 3 − 7 ( − 2 ) 2 − 10 ( − 2 ) + 6 = 32 − 40 − 28 + 20 + 6 = − 10 ∴ f ( − 2 ) = − 10

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ISBN: 9781285965918

Author: Gilbert

Publisher: Cengage

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Chapter 8.4, Problem 15E

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Factor of the polynomial 2x4+5x3−7x2−10x+6 as a product of its leading coefficient and a finite number of monic irreducible polynomial over the field of rational numbers.

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Chapter 8.4, Problem 14E

Chapter 8.4, Problem 16E

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Factor of the polynomial 2 x 4 + 5 x 3 − 7 x 2 − 10 x + 6 as a product of its leading coefficient and a finite number of monic irreducible polynomial over the field of rational numbers.

Solution Summary: The author explains the formula for determining the factor of the polynomial 2x4+5.

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Factor of the polynomial 2x4+5x3−7x2−10x+6 as a product of its leading coefficient and a finite number of monic irreducible polynomial over the field of rational numbers.

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Chapter 8 Solutions