## Precalculus (6th Edition)

Published by Pearson

# Chapter R - Review of Basic Concepts - R.4 Factoring Polynomials - R.4 Exercises - Page 44: 102

#### Answer

$(2p^{2}+3)(3p^{2}-1)$

#### Work Step by Step

Substitute $u=p^{2}$ $6p^{4}+7p^{2}-3=6u^{2}+7u-3=...$ 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 $-2$ $6u^{2}+7u-3=6u^{2}+9u-2u-3=$ $=3u(2u+3)-(2u+3)$ $=(2u+3)(3u-1)$ ... bring back p $=(2p^{2}+3)(3p^{2}-1)$

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