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: 18

Answer

$f(x)= (x-1)(x-2)(2x+3)$

Work Step by Step

If k is a zero of f, then, by the factor theorem, (x-k) is a factor of f(x). Synthetic division gives us a quotient q(x), with degree one less than f(x). and we know the remainder is zero, so $f(x)=(x-k)\cdot q(x).$ We use this first. $k=1$ $\left\{\begin{array}{lllllll} 1 & | & 2 & -3 & -5 & 6 & \\ & & & 2 & -1 & -6 & \\ & & -- & -- & -- & -- & --\\ & & 2 & -1 & -6 & \fbox{$0$} & \end{array}\right.$ $f(x)= (x-1)(2x^{2}-x-6)$ We now factor the trinomial $ax^{2}+bx+c$ by searching for two factors of $ac$ whose sum is $b.$ If they exist, rewrite $bx$ and factor in pairs. Factors of $2(-6)=-12$ whose sum is $-1$ ... are $-4$ and $+3$: $2x^{2}-x-6=2x^{2}-4x+3x-6$ $=2x(x-2)+3(x-2)=(x-2)(2x+3)$ So, $f(x)= (x-1)(x-2)(2x+3)$
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