#### Answer

$2m(4m+3)(3m-4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
24m^3-14m^2-24m
,$ 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.
$\bf{\text{Solution Details:}}$
The $GCF$ of the constants of the terms $\{
24,-14,-24
\}$ is $
2
$ since it is the highest number that can divide all the given constants. The $GCF$ of the common variable/s is the variable/s with the lowest exponent. Hence, the $GCF$ of the common variable/s $\{
m^3,m^2,m
\}$ is $
m
.$ Hence, the entire expression has $GCF=
2m
.$
Factoring the $GCF=
2m
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
2m \left( \dfrac{24m^3}{2m}-\dfrac{14m^2}{2m}-\dfrac{24m}{2m}
\right)
\\\\=
2m \left( 12m^2-7m-12 \right)
.\end{array}
In the trinomial expression above the value of $ac$ is $
12(-12)=-144
$ and the value of $b$ is $
-7
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
9,-16
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2m \left( 12m^2+9m-16m-12 \right)
.\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}
2m [(12m^2+9m)-(16m+12)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2m [3m(4m+3)-4(4m+3)]
.\end{array}
Factoring the $GCF=
(4m+3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2m [(4m+3)(3m-4)]
\\\\=
2m(4m+3)(3m-4)
.\end{array}