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: 56

Answer

a. $\pm1,\pm3,\pm5,\pm15,\pm1/2,\pm3/2,\pm5/2,\pm15/2,\pm1/4,\pm3/4,\pm5/4,\pm15/4,\pm1/8,\pm3/8,\pm5/8,\pm15/8$ b. positive real zeros: $3,1$. negative real zeros: none. c. $x=\frac{1}{2}$ see figure. d. $x=\frac{3}{2}, \frac{5}{2}$ (and $x=1/2$ from c).
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Work Step by Step

a. Given the function $f(x)=8x^3-36x^2+46x-15$ we have $p=\pm1,\pm3,\pm5,\pm15, q=\pm1,\pm2,\pm4,\pm8$ and the possible rational zeros are $\frac{p}{q}=\pm1,\pm3,\pm5,\pm15,\pm1/2,\pm3/2,\pm5/2,\pm15/2,\pm1/4,\pm3/4,\pm5/4,\pm15/4,\pm1/8,\pm3/8,\pm5/8,\pm15/8$ b. There are 3 sign changes in $f(x)$ and based on Descartes’s Rule of Signs, we can determine the possible number of positive real zeros to be: $3,1$. We examine $f(-x)=8(-x)^3-36(-x)^2+46(-x)-15=-8x^3-36x^2-46x-15$ We can find 0 sign change and based on Descartes’s Rule of Signs, we can determine that there are 0 negative real zeros. c. We can use synthetic division to find a zero as $x=\frac{1}{2}$, as shown in the figure. d. Part (c) gives a quotient $8x^2-32x+30=2(4x^2-16x+15)=2(2x-3)(2x-5)$; thus the other two zeros are $x=\frac{3}{2}, \frac{5}{2}$ (and $x=1/2$ from c).
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