Answer
$5(6y-5)(2y+9)$
Work Step by Step
Factoring the $GCF=
5
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
220y+60y^2-225
\\\\=
60y^2+220y-225
\\\\=
5(12y^2+44y-45)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
5(12y^2+44y-45)
\end{array} has $ac=
12(-45)=-540
$ and $b=
44
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-10,54
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5(12y^2-10y+54y-45)
.\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}
5[(12y^2-10y)+(54y-45)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5[2y(6y-5)+9(6y-5)]
.\end{array}
Factoring the $GCF=
(6y-5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
5[(6y-5)(2y+9)]
\\\\=
5(6y-5)(2y+9)
.\end{array}
