Answer
$-3a(2b+5)(b-15)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
-6ab^2+75ab-225a
,$ factor first the negative $GCF.$ Then 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.
$\bf{\text{Solution Details:}}$
Factoring the negative $GCF$ equal $
-3a
,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
-3a(2b^2-25b-75)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
2(-75)=-150
$ and the value of $b$ is $
-25
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
5,-30
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
-3a(2b^2+5b-30b-75)
.\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}
-3a[(2b^2+5b)-(30b+75)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-3a[b(2b+5)-15(2b+5)]
.\end{array}
Factoring the $GCF=
(2b+5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
-3a[(2b+5)(b-15)]
\\\\=
-3a(2b+5)(b-15)
.\end{array}
