Answer
$-5(3a-4)(a+6)$
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 $GCF=
-5
,$ the given expression, $
-15a^2-70a+120
,$ is equivalent to
\begin{array}{l}\require{cancel}
-5(3a^2+14a-24)
.\end{array}
In the expression above, the value of $c$ is $
3(-24)=-72
$ and the value of $b$ is $
14
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-72\}, \{2,-36\}, \{3,-24\}, \{4,-18\}, \{8,-9\},
\{-1,72\}, \{-2,36\}, \{-3,24\}, \{-4,18\}, \{-8,9\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-4,18
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
-5(3a^2-4a+18a-24)
.\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}
-5[(3a^2-4a)+(18a-24)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-5[a(3a-4)+6(3a-4)]
.\end{array}
Factoring the $GCF=
(3a-4)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-5[(3a-4)(a+6)]
\\\\=
-5(3a-4)(a+6)
.\end{array}