Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - Test: 15

Answer

$\color{blue}{(2x-1)(3x-7)}$

Work Step by Step

RECALL: A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$ If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping. The given trinomial h as $a=6$, $b=-17$, and $c=7$. Thus, $ac = 6(7) = 42$ Note that $42=-3(-14)$ and $-3+(-14)=-17$ This means that $d=-3$ and $e=-14$. Rewrite the middle term of the trinomial as $-3x$ +$(-14x)$ to obtain: $6x^2-17x+7 =6x^2-3x-14x+7 $ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(6x^2-3x)+(-14x+7) \\=3x(2x-1) +(-7)(2x-1)$ Factor out the GCF $2x-1$ to obtain: $=(2x-1)(3x+(-7)) \\=\color{blue}{(2x-1)(3x-7)}$
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