Answer
$-2(2a-3)(3a+7)$
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 negative $GCF=
-2
,$ the given expression, $
-12a^2-10a+42
,$ is equivalent to
\begin{array}{l}\require{cancel}
-2(6a^2+5a-21)
.\end{array}
In the expression above, the value of $c$ is $
6(-21)=-126
$ and the value of $b$ is $
5
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-126\}, \{2,-63\}, \{3,-42\}, \{6,-21\}, \{9,-14\},
\{-1,126\}, \{-2,63\}, \{-3,42\}, \{-6,21\}, \{-9,14\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-9,14
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
-2(6a^2-9a+14a-21)
.\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[(6a^2-9a)+(14a-21)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-2[3a(2a-3)+7(2a-3)]
.\end{array}
Factoring the $GCF=
(2a-3)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-2[(2a-3)(3a+7)]
\\\\=
-2(2a-3)(3a+7)
.\end{array}
