Explanation Given Information: The provided polynomial is x 3 + 2 x 2 − 5 x − 6 . Formula used: Rational Zeros Theorem: If a polynomial ( a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a o ) has integer coefficient, then every rational zero is of the form x = p q , where p is a factor of a o , and q is a factor of a n . Calculation: Consider the polynomial, p ( x ) = x 3 + 2 x 2 − 5 x − 6 Factor of constant term ( − 6 ) of polynomial x 3 + 2 x 2 − 5 x − 6 will be ± 1 , ± 2 ± 3 and ± 6 . Factor of leading coefficient ( 1 ) will be ± 1 . Now apply, Rational Zeros Theorem for polynomial p ( x ) to find zeros. The possible rational zeros are the factor of the constant term divided by the factors of the leading coefficient that is, ± 1 , ± 2 ± 3 and ± 6 The possible rational factor of polynomial x 3 + 2 x 2 − 5 x − 6 is ± 1 , ± 2 ± 3 and ± 6 . Now, apply synthetic devisor method for x = 2 . 2 1 2 − 5 − 6 2 8 6 1 4 3 0 The factor of polynomial x 3 + 2 x 2 − 5 x − 6 are ( x − 2 ) ( x 2 + 4 x + 3 ) . Consider the quadratic equation. x 2 + 4 x + 3 Factor the quadratic equation x 2 + 4 x + 3 x 2 + x − 6 , ( x 2 + 4 x + 3 ) = x 2 + 3 x + x + 3 = x ( x + 3 ) + 1 ( x + 3 ) = ( x + 1 ) ( x + 3 ) The factor of polynomial x 3 + 2 x 2 − 5 x − 6 are ( x + 1 ) ( x − 2 ) ( x + 3 ) . To find zeros set first term of the factor ( x + 1 ) ( x − 2 ) ( x + 3 ) equal to zero

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ISBN: 9780357265161

Author: Larson

Publisher: CENGAGE L

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Chapter A4, Problem 68E

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Chapter A4, Problem 67E

Chapter A4, Problem 69E

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Chapter A4 Solutions

CALCULUS: APPLIED APPROACH LOW COST MA

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The factor the polynomial x 3 + 2 x 2 − 5 x − 6 .

Solution Summary: The author explains the rational zeros of the polynomial p(x).

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Chapter A4 Solutions