Intermediate Algebra (6th Edition)

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

Chapter 5 - Section 5.7 - Factoring by Special Products - Exercise Set: 9

Answer

$x^2(3y+2)^2$

Work Step by Step

Factoring the $GCF=x^2$, then the given expression, $ 9y^2x^2+12yx^2+4x^2 $, is equivalent to \begin{array}{l} x^2(9y^2+12y+4) .\end{array} The two numbers whose product is $ac= 9(4)=36 $ and whose sum is $b= 12 $ are $\{ 6,6 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ x^2(9y^2+12y+4) $, is \begin{array}{l}\require{cancel} x^2(9y^2+6y+6y+4) \\\\= x^2[(9y^2+6y)+(6y+4)] \\\\= x^2[3y(3y+2)+2(3y+2)] \\\\= x^2[(3y+2)(3y+2)] \\\\= x^2(3y+2)^2 .\end{array}
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