Answer
$5wxz(w^2x+8z)(w^2x-3z)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
5w^5x^3z+25w^3x^2z^2-120wxz^3
,$ factor first the $GCF.$ 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.
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms is $GCF=
5wxz
$ since it is the highest expression that can evenly divide (no remainder) all the given terms. Factoring the $GCF,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
5wxz(w^4x^2+5w^2xz-24z^2)
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
1(-24)=-24
$ and the value of $b$ is $
5
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
8,-3
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
5wxz(w^4x^2+8w^2xz-3w^2xz-24z^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}
5wxz[(w^4x^2+8w^2xz)-(3w^2xz+24z^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
5wxz[w^2x(w^2x+8z)-3z(w^2x+8z)]
.\end{array}
Factoring the $GCF=
(w^2x+8z)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
5wxz[(w^2x+8z)(w^2x-3z)]
\\\\=
5wxz(w^2x+8z)(w^2x-3z)
.\end{array}
