#### Answer

$a(4a-3)(3a+1)$

#### Work Step by Step

Factoring the $GCF=a,$ then
\begin{array}{l}\require{cancel}
12a^3-5a^2-3a
\\\\=
a(12a^2-5a-3)
.\end{array}
Using the factoring of trinomials in the form $ax^2+bx+c,$, the value of $ac$ in the trinomial expression above is $
12(-3)=-36
$ and the value of $b$ is $
-5
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
-9,4
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
a(12a^2-9a+4a-3)
.\end{array}
Grouping the first and second terms and the third and fourth terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
a[(12a^2-9a)+(4a-3)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
a[3a(4a-3)+(4a-3)]
.\end{array}
Factoring the $GCF=
(4a-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
a[(4a-3)(3a+1)]
\\\\=
a(4a-3)(3a+1)
.\end{array}