Answer
$2(2y+9)(8y-3)$
Work Step by Step
Factoring the $GCF=
2
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
132y+32y^2-54
\\\\=
32y^2+132y-54
\\\\=
2(16y^2+66y-27)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
2(16y^2+66y-27)
\end{array} has $ac=
16(-27)=-432
$ and $b=
66
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
72,-6
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2(16y^2+72y-6y-27)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
2[(16y^2+72y)-(6y+27)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2[8y(2y+9)-3(2y+9)]
.\end{array}
Factoring the $GCF=
(2y+9)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2[(2y+9)(8y-3)]
\\\\=
2(2y+9)(8y-3)
.\end{array}