Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 84



Work Step by Step

Substitute $u=4z-3$ $6u^{2}+7u-3=...$ When factoring $ax^{2}+bx+c$, we search for factors of $ac$ whose sum is $b,$ and, if we find them, we rewrite $bx$ and proceed to factor in groups. Here, factors of $(6)\times(-3)=-18$ that add to $+7$ are ....$+9$ and $-2$ $...=6u^{2}+9u-2u-3$ $=3u(2u+3)-(2u+3)$ $=(2u+3)(3u-1)$ ... bring back z $=[2(4z-3)+3][3(4z-3)-1]$ $=(8z-6+3)(12z-9-1)$ $=(8z-3)(12z-10)$ ... factor out $2$ from the second parentheses $=2(8z-3)(6z-5)$
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