#### Answer

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

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
24x-2x^2-2x^3
,$ factor first the $GCF.$ Then factor the resulting trinomial by finding 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 $\{
24,-2,-2
\}$ 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 $\{
x,x^2,x^3
\}$ is $
x
.$ Hence, the entire expression has $GCF=
2x
.$
Factoring the $GCF=
2x
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
2x \left( \dfrac{24x}{2x}-\dfrac{2x^2}{2x}-\dfrac{2x^3}{2x}
\right)
\\\\=
2x (12-x-x^2)
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
-1(12)=-12
$ and the value of $b$ is $
-1
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-12\}, \{2,-6\}, \{3,-4\},
\\
\{-1,12\}, \{-2,6\}, \{-3,4\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
3,-4
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
2x (12+3x-4x-x^2)
.\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}
2x [(12+3x)-(4x+x^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2x [3(4+x)-x(4+x)]
.\end{array}
Factoring the $GCF=
(4+x)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2x [(4+x)(3-x)]
\\\\=
2x (4+x)(3-x)
.\end{array}