College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Polynomial and Rational Functions - Exercise Set 3.4 - Page 389: 86

Answer

Makes sense.

Work Step by Step

The Rational Zero Theorem states that if $\displaystyle \frac{p}{q}$ (where $p$ is reduced to lowest terms) is a rational zero of $f$, then $p$ is a factor of the constant term, $a_{0}$, and $q$ is a factor of the leading coefficient, $a_{n}$. -------------- 1. So, if there are rational zeros $\displaystyle \frac{p}{q}$, then $p;\quad \pm 1, \pm 2$ $q:\quad \pm 1$ possible $\displaystyle \frac{p}{q}:\quad \pm 1, \pm 2.$ There are four, so the first part of the statement is true. 2. The second part can be checked in several ways: Evaluate f(possible root) ... see if any yield 0, test the possible $\displaystyle \frac{p}{q}$ with synthetic division, graph and analyze - see if there are any intersections with the x-axis, or, like we do here, algebraically conclude that, for real x: 1. $x^{4}$ and $x^{2}$ are never negative (even exponents), 2. $x^{4}+3x^{2}+2$ is never less than 2, so 3. f(x) is never zero ... f has no real roots. Verdict: makes sense.
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