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: 59

Answer

a. $\frac{p}{q}=\pm1,\pm2,\pm1/2,\pm1/4$ b. $1$, $1$; see explanations. c. $x=\pm\frac{1}{2}$ d. $x=\pm\sqrt 2i$ (and the zeros from c).
1582639897

Work Step by Step

a. Given the equation $4x^4+7x^2-2=0$ we have $p=\pm1,\pm2, q=\pm1,\pm2,\pm4$ and the possible rational zeros (roots) are $\frac{p}{q}=\pm1,\pm2,\pm1/2,\pm1/4$ b. There is 1 sign change in the equation and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be $1$ We examine $f(-x)=4(-x)^4+7(-x)^2-2=4x^4+7x^2-2$ We can find 1 sign change and based on Descartes’s Rule of Signs, we can determine that the number of negative real zeros is $1$. c. We can use synthetic division to find two roots as $x=\pm\frac{1}{2}$, as shown in the figure. d. Part (c) gives a quotient of $4x^2+8=4(x^2+2)$; thus the other two roots are $x=\pm\sqrt 2i$ (and the zeros from c).
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.