#### Answer

$2(7y+2)(2y+1)$

#### Work Step by Step

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}