#### Answer

$4m(2n+3)(5n-7)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
40mn^2+4mn-84m
,$ factor first the $GCF.$ Then find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
4m
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
4m(10n^2+n-21)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
10(-21)=-210
$ and the value of $b$ is $
1
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
15,-14
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
4m(10n^2+15n-14n-21)
.\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}
4m[(10n^2+15n)-(14n+21)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
4m[5n(2n+3)-7(2n+3)]
.\end{array}
Factoring the $GCF=
(2n+3)d
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
4m[(2n+3)(5n-7)]
\\\\=
4m(2n+3)(5n-7)
.\end{array}