## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 303: 59

#### Answer

$3(a+2b)^2$

#### Work Step by Step

Factoring the $GCF=3$, then the given expression, $3a^2+12ab+12b^2$, is equivalent to $3(a^2+4ab+4b^2)$.\\ The two numbers whose product is $ac= 1(4)=4$ and whose sum is $b= 4$ are $\{ 2,2 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $3(a^2+4ab+4b^2)$, is \begin{array}{l}\require{cancel} 3(a^2+2ab+2ab+4b^2) \\\\= 3[(a^2+2ab)+(2ab+4b^2)] \\\\= 3[a(a+2b)+2b(a+2b)] \\\\= 3[(a+2b)(a+2b)] \\\\= 3(a+2b)^2 .\end{array}

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