Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - Summary Exercises on Polynomial Functions, Zeros, and Graphs - Exercises - Page 359: 5

Answer

(a) 3 or 1 positive real zeros. 1 negative real zero. 0 or 2 complex zeros. (b) $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{1}{6}$ (c) $x=\frac{1}{2},\frac{1}{3}$ (d) $x=\pm\sqrt 2$

Work Step by Step

Given $f(x)=6x^4-5x^3-11x^2+10x-2$ (a) For $f(x)$, we can identify 3 sign changes. Based on the Descartes’ rule of signs, we can determine that there are 3 or 1 positive real zeros. For $f(-x)$, we have $f(-x)=6x^4+5x^3-11x^2-10x-2$, and we can identify 1 sign change, thus there is 1 negative real zero. Since the degree of order is $4$, there are 0 or 2 complex zeros. (b) Based on the rational zeros theorem, we have $p=\pm1,\pm2$ and $q=\pm1,\pm2,\pm3,\pm6$, thus we can list the possible rational zeros as $\frac{p}{q}=\pm1,\pm2,\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{2}{3},\pm\frac{1}{6}$ (c) Use synthetic division as shown in the figure, we can find a zero as $x=\frac{1}{2},\frac{1}{3}$ and factor the function as $f(x)=(x-\frac{1}{2})(x-\frac{1}{3})(6x^2-12)$. Thus, we can find the rational zeros as $x=\frac{1}{2},\frac{1}{3}$ (d) We can find other complex zeros as $x^2=2$ and $x=\pm\sqrt 2$
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