Precalculus (6th Edition) Blitzer

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

Chapter 2 - Review Exercises - Page 431: 60

Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm4,\pm1/2,$ b. positive real zeros: $2,0$, negative real zeros: $2,0$. c. $x=\pm2$ d. $x=-\frac{1}{2},1$ (and the zeros from c).
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Work Step by Step

a. Given the equation $2x^4+x^3-9x^2-4x+4=0$ we have $p=\pm1,\pm2,\pm4, q=\pm1,\pm2$ and the possible rational zeros (roots) are $\frac{p}{q}=\pm1,\pm2,\pm4,\pm1/2$ b. There are 2 sign changes in the equation and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be: $2,0$. We examine $f(-x)=2(-x)^4+(-x)^3-9(-x)^2-4(-x)+4=2x^4-x^3-9x^2+4x+4$ We can find 2 sign changes and based on Descartes’s Rule of Signs, we can determine that the number of negative real zeros is: $2,0$. c. We can use synthetic division to find two roots as $x=\pm2$, as shown in the figure. d. Part (c) gives a quotient of $2x^2+x-1=(2x+1)(x-1)$; thus the other two roots are $x=-\frac{1}{2},1$ (and the zeros from c).
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