Answer
$(3r+1)(4r-3)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
12r^2-5r-3
,$ 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.
$\bf{\text{Solution Details:}}$
To factor the trinomial expression above, note that the value of $ac$ is $
12(-3)=-36
$ and the value of $b$ is $
-5
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-36\}, \{2,-18\}, \{3,-12\}, \{4,-9\}, \{6,-6\},
\\
\{-1,36\}, \{-2,18\}, \{-3,12\}, \{-4,9\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
4,-9
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
12r^2+4r-9r-3
.\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}
(12r^2+4r)-(9r+3)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
4r(3r+1)-3(3r+1)
.\end{array}
Factoring the $GCF=
(3r+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(3r+1)(4r-3)
.\end{array}