#### Answer

$(6xy-5)(3xy+2)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
18x^2y^2-3xy-10
\end{array} has $ac=
18(-10)=-180
$ and $b=
-3
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-15,12
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
18x^2y^2-15xy+12xy-10
.\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}
(18x^2y^2-15xy)+(12xy-10)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
3xy(6xy-5)+2(6xy-5)
.\end{array}
Factoring the $GCF=
(6xy-5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(6xy-5)(3xy+2)
.\end{array}