Answer
$4pq(3p+5q)(2p+q)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
24p^3q+52p^2q^2+20pq^3
,$ factor first the $GCF.$ Then use the factoring of trinomials in the form $ax^2+bx+c.$
$\bf{\text{Solution Details:}}$
The $GCF$ of the terms in the given expression is $
4pq
,$ since it is the greatest expression that can divide all the terms evenly (no remainder.) Factoring the $GCF$ results to
\begin{array}{l}\require{cancel}
4pq(6p^2+13pq+5q^2)
.\end{array}
In the trinomial expression above, $a=
6
,b=
13
,\text{ and } c=
5
.$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac=
6(5)=30
$ and whose sum is $b$ are $\left\{
10,3
\right\}.$ Using these two numbers to decompose the middle term results to
\begin{array}{l}\require{cancel}
4pq(6p^2+10pq+3pq+5q^2)
.\end{array}
Using factoring by grouping, the expression above is equivalent to
\begin{array}{l}\require{cancel}
4pq[(6p^2+10pq)+(3pq+5q^2)]
\\\\=
4pq[2p(3p+5q)+q(3p+5q)]
\\\\=
4pq[(3p+5q)(2p+q)]
\\\\=
4pq(3p+5q)(2p+q)
.\end{array}
