College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 3 - Section 3.3 - Zeros of Polynomial Functions - 3.3 Exercises - Page 316: 51

Answer

$f(x)=-\displaystyle \frac{1}{2}x^{3}-\frac{1}{2}x^{2}+x$

Work Step by Step

For any polynomial function $f(x),\ (x-k)$ is a factor of the polynomial if and only if $f(k)=0$. So, we can write $ f(x)=a(x+2)(x-1)(x-0)\quad$ for some number a. To find $a,$ use the given information: $f(2)=-1$ $f(2)=a(-1+2)(-1-1)(-1-0)=-1$ $a(1)(-2)(-1)=-1$ $2a=-1$ $a=-1/2$ So, $f(x)=-\displaystyle \frac{1}{2}(x+2)(x-1)(x-0)$ Rewrite in standard form $f(x)=-\displaystyle \frac{1}{2}x(x^{2}+x-2)$ $f(x)=-\displaystyle \frac{1}{2}x^{3}-\frac{1}{2}x^{2}+x$
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