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 6 y 3 + 19 y 2 + 2 y − 3 = 0 . Here, f ( y ) = 6 y 3 + 19 y 2 + 2 y − 3 and the degree of the polynomial is 3. Move from left to right, obtain the number of sign changes. From 2 y → − 3 , there is only 1 sign changes. Thus, the number of positive roots is 1. Change the function f ( y ) to f ( − y ) as follows. f ( − y ) = 6 ( − y ) 3 + 19 ( − y ) 2 + 2 ( − y ) − 3 = − 6 y 3 + 19 y 2 − 2 y − 3 From − 6 y 3 → 19 y 2 and − 2 y → − 3 , there are 2 sign changes. Thus, the number of negative roots is 2. Hence, the maximum possible numbers of positive and negative roots of 6 y 3 + 19 y 2 + 2 y − 3 = 0 are 1 and 2, respectively. Thus, the possible rational roots by Rational Root Theorem are as follows. r r = integer factor of − 3 integer factor of 6 = ± 1 , ± 3 ± 1 , ± 2 , ± 3 , ± 6 = ± 1 , ± 3 , ± 1 2 , ± 3 2 , ± 1 3 Begin with − 1 , 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 42RE

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Chapter 15, Problem 41RE

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Chapter 15 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 15.1 - Use the factor theorem to determine whether t − 2...Ch. 15.1 - Prob. 2PE Ch. 15.1 - Prob. 3PE Ch. 15.1 - Prob. 4PE Ch. 15.1 - Prob. 1E Ch. 15.1 - Prob. 2E Ch. 15.1 - In Exercises 1–4, make the given changes in the...Ch. 15.1 - Prob. 4E Ch. 15.1 - Prob. 5E Ch. 15.1 - Prob. 6E

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The equation 6 y 3 + 19 y 2 + 2 y − 3 = 0 .

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Chapter 15 Solutions