#### Answer

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

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{
expression
}$
\begin{array}{l}\require{cancel}
12x^2-x-1
\end{array} has $ac=
12(-1)=-12
$ and $b=
-1
.$
The two numbers with a product of $c$ and a sum of $b$ are $\left\{
-4,3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
12x^2-4x+3x-1
.\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}
(12x^2-4x)+(3x-1)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
4x(3x-1)+(3x-1)
.\end{array}
Factoring the $GCF=
(3x-1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3x-1)(4x+1)
.\end{array}