Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set: 59



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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