Answer
$2(6x-1)(3x+2)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, factor the GCF of the terms. Then, to factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
Using the $GCF=
2
,$ the given expression, $
36x^2+18x-4
,$ is equivalent to
\begin{array}{l}\require{cancel}
2(18x^2+9x-2)
.\end{array}
In the expression above, the value of $ac$ is $
18(-2)=-36
$ and the value of $b$ is $
9
.$
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\}, \{-6,6\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-3,12
\}.$ Using these $2$ numbers to decompose the middle term of the expression above results to
\begin{array}{l}\require{cancel}
2(18x^2-3x+12x-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}
2[(18x^2-3x)+(12x-2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
2[3x(6x-1)+2(6x-1)]
.\end{array}
Factoring the $GCF=
(6x-1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
2[(6x-1)(3x+2)]
\\\\=
2(6x-1)(3x+2)
.\end{array}
