Explanation Theorem used: Rational Root Theorem: Any rational root r r of a polynomial equation f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 0 = 0 is an integer factor of a 0 divided by an integer factor of a n . Therefore, r r = integer factor of a 0 integer factor of a n . Result used: Descartes’ Rule of Signs: “(a) The number of positive zeros of a polynomial equation f ( x ) = 0 cannot exceed than the number of changes in sign in f ( x ) in going from one term to the next in f ( x ) . (b) The number of negative zeros cannot exceed the number of sign changes in f ( − x ) ”. Calculation: It is given that the polynomial equation is 2 x 4 + 5 x 3 − 14 x 2 − 23 x + 30 = 0 . Here, f ( x ) = 2 x 4 + 5 x 3 − 14 x 2 − 23 x + 30 and the degree is 4. Move from left to right, obtain the number of sign changes. From 5 x 3 → − 14 x 2 and − 23 x → 30 , there are 2 sign changes. Thus, the number of positive roots is 2. Change the function f ( x ) to f ( − x ) as follows. f ( − x ) = 2 ( − x ) 4 + 5 ( − x ) 3 − 14 ( − x ) 2 − 23 ( − x ) + 30 = 2 x 4 − 5 x 3 − 14 x 2 + 23 x + 30 Move from left to right, obtain the number of sign changes. From 2 x 4 → − 5 x 3 and − 14 x 2 → − 23 x , there are 2 sign changes. Hence, the maximum possible numbers of positive and negative roots of 2 x 4 + 5 x 3 − 14 x 2 − 23 x + 30 = 0 are 2 and 2, respectively. Thus, the possible rational roots by Rational Root Theorem are, r r = integer factor of 30 integer factor of 2 = ± 1 , ± 2 , ± 3 , ± 5 , ± 6 , ± 10 , ± 15 ± 1 , ± 2 = ± 1 , ± 2 , ± 3 , ± 5 , ± 6 , ± 10 , ± 15 , ± 1 2 , ± 3 2 , ± 5 2 , ± 15 2 Begin with 1 2 , perform synthetic division as follows

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Chapter 15, Problem 44RE

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Chapter 15, Problem 43RE

Chapter 15, Problem 45RE

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Chapter 15 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

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The equation 2 x 4 + 5 x 3 − 14 x 2 − 23 x + 30 = 0 .

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Chapter 15 Solutions