Answer
$3mn(3m+2n)(2m-n)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
18m^3n+3m^2n^2-6mn^3
,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $ax^2+bx+c.$
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms in the given expression is $
3mn
,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to
\begin{array}{l}\require{cancel}
3mn(6m^2+mn-2n^2)
.\end{array}
In the trinomial expression above, $a=
6
,b=
1
,\text{ and } c=
-2
.$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac=
6(-2)=-12
$ and whose sum is $b$ are $\left\{
4,-3
\right\}.$ Using these two numbers to decompose the middle term results to
\begin{array}{l}\require{cancel}
3mn(6m^2+4mn-3mn-2n^2)
.\end{array}
Using factoring by grouping, the expression above is equivalent to
\begin{array}{l}\require{cancel}
3mn[(6m^2+4mn)-(3mn+2n^2)]
\\\\=
3mn[2m(3m+2n)-n(3m+2n)]
\\\\=
3mn[(3m+2n)(2m-n)]
\\\\=
3mn(3m+2n)(2m-n)
.\end{array}
