## Intermediate Algebra (6th Edition)

$2(7y+2)(2y+1)$
Factoring the $GCF=2$, then the given expression, $28y^2+22y+4$, is equivalent to \begin{array}{l}\require{cancel} 2(14y^2+11y+2) .\end{array} The two numbers whose product is $ac= 14(2)=28$ and whose sum is $b= 11$ are $\{ 4,7 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $2(14y^2+11y+2)$, is \begin{array}{l}\require{cancel} 2(14y^2+4y+7y+2) \\\\= 2[(14y^2+4y)+(7y+2)] \\\\= 2[2y(7y+2)+(7y+2)] \\\\= 2[(7y+2)(2y+1)] \\\\= 2(7y+2)(2y+1) .\end{array}