## Intermediate Algebra (6th Edition)

$2(x+4y)^2$
Factoring the $GCF=2$, then the given expression, $2x^2+16xy+32y^2$, is equivalent to $2(x^2+8xy+16y^2)$.\\ The two numbers whose product is $ac= 1(16)=16$ and whose sum is $b= 8$ are $\{ 4,4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $2(x^2+8xy+16y^2)$, is \begin{array}{l}\require{cancel} 2(x^2+4xy+4xy+16y^2) \\\\= 2[(x^2+4xy)+(4xy+16y^2)] \\\\= 2[x(x+4y)+4y(x+4y)] \\\\= 2[(x+4y)(x+4y)] \\\\= 2(x+4y)^2 .\end{array}