Explanation Given Information: The provided polynomial is ( x 3 − 6 x 2 + 11 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 − 6 x 2 + 11 x − 6 ) Factor of constant term ( − 6 ) of polynomial x 3 − 6 x 2 + 11 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 − 6 x 2 + 11 x − 6 is ± 1 , ± 2 ± 3 and ± 6 . Now, apply synthetic devisor method for x = 1 . 1 1 − 6 11 − 6 1 − 5 6 1 − 5 6 0 The factor of polynomial x 3 − 6 x 2 + 11 x − 6 are ( x − 1 ) ( x 2 − 5 x + 6 ) . Consider the quadratic equation. ( x 2 − 5 x + 6 ) Factor the quadratic equation ( x 2 − 5 x + 6 ) , ( x 2 − 5 x + 6 ) = x 2 − 3 x − 2 x + 6 = x ( x − 3 ) − 2 ( x − 3 ) = ( x − 2 ) ( x − 3 ) The factor of polynomial ( x 3 − 6 x 2 + 11 x − 6 ) are ( x − 1 ) ( x − 2 ) ( x − 3 )

Calculus: An Applied Approach (MindTap Course List)

10th Edition

ISBN: 9781305860919

Author: Ron Larson

Publisher: Cengage Learning

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Chapter A4, Problem 67E

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Chapter A4, Problem 66E

Chapter A4, Problem 68E

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Calculus: An Applied Approach (MindTap Course List)

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The real zeros of polynomial ( x 3 − 6 x 2 + 11 x − 6 ) .

Solution Summary: The author calculates the real zeros of the polynomial (x3-62+11x 6).

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Chapter A4 Solutions