Answer
$3(5y^2-4)(y^2+3)$
Work Step by Step
Factoring the $GCF=3,$ the given expression, $
15y^4+33y^2-36
,$ is equivalent to
\begin{align*}
3(5y^4+11y^2-12)
.\end{align*}
Using the factoring of trinomials in the form $ax^2+bx+c,$ the expression
\begin{align*}
(5y^4+11y^2-12)
\end{align*} has $ac=
5(-12)=-60
$ and $b=
11
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-4,15
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{align*}
3(5y^4-4y^2+15y^2-12)
.\end{align*}
Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to
\begin{align*}
3[(5y^4-4y^2)+(15y^2-12)]
.\end{align*}
Factoring the $GCF$ in each group results to
\begin{align*}
3[y^2(5y^2-4)+3(5y^2-4)]
.\end{align*}
Factoring the $GCF=
(5y^2-4)
$ of the entire expression above results to
\begin{align*}
&
3[(5y^2-4)(y^2+3)]
\\&=
3(5y^2-4)(y^2+3)
.\end{align*}
Hence the factored form of $15y^4+33y^2-36$ is $3(5y^2-4)(y^2+3)$.
