Answer
$6(a+4)(3a-8)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
18a^2+24a-192
,$ 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 $GCF=
6
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
6(3a^2+4a-32)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
3(-32)=-96
$ and the value of $b$ is $
4
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
12,-8
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
6(3a^2+12a-8a-32)
.\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}
6[(3a^2+12a)-(8a+32)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
6[3a(a+4)-8(a+4)]
.\end{array}
Factoring the $GCF=
(a+4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
6[(a+4)(3a-8)]
\\\\=
6(a+4)(3a-8)
.\end{array}
