Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.5 - Zeros of Polynomial Functions - Exercise Set - Page 379: 68

Answer

The possible number of negative roots of a polynomial equation can be determined by calculating the number of sign changes using Descartes’s Rule of Signs.

Work Step by Step

To find possibilities for negative real zeros, count the number of sign changes in the equation for f (-x). We obtain this equation by replacing x with -x in the given function. For example, consider the polynomial function\[f\left( x \right)={{x}^{3}}+2{{x}^{2}}+5x+4\]…… (1) Replace x with –x in equation (1). \[\begin{align} & f\left( -x \right)={{\left( -x \right)}^{3}}+2{{\left( -x \right)}^{2}}+5\left( -x \right)+4 \\ & f\left( -x \right)=-{{x}^{3}}+2{{x}^{2}}-5x+4 \\ \end{align}\] The equation with \[f\left( -x \right)\]is \[f\left( -x \right)=-{{x}^{3}}+2{{x}^{2}}-5x+4\] ….. (2) Now, we see that there are three variations in sign. The number of negative real zeros of f (x) is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that either there are 3 negative real zeros or there is 3 - 2 = 1 negative real zero.
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