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 315: 7

Answer

Not a factor.

Work Step by Step

Using synthetic division =, we divide $f(x)$ by $x-k$ to obtain the quotient $q(x)$ and remainder $r(x)$ such that $f(x)=(x-k)\cdot q(x)+r(x)$ $k=-1$ $\left\{\begin{array}{llllllll} -1 & | & 2 & 5 & -8 & 3 & 13 & \\ & & & -2 & -3 & 11 & -14 & \\ & & -- & -- & -- & -- & -- & \\ & & 2 & 3 & -11 & 14 & \fbox{$-1$} & \end{array}\right.$ The last row gives us the coefficients of the quotient and the remainder (boxed). By the remainder theorem, $f(-1)=-1$, so $k=-1$ is not a zero of f. Apply the factor theorem, which states that the following statements : $\left\{\begin{array}{l} \text{$k$ is a zero of f}\\ \text{$(x-k)$ is a factor of f} \end{array}\right.\quad $ are equivalent. Since $-1$ is not a zero of f, $(x-k)$ = $(x+1)$ is not a factor of f.
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