Answer
$13y(y+1)(y-4)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
First, factor the GCF of the terms. Then, to factor the quadratic expression $x^2+bx+c,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$. Express the factored form as $GCF(x+m_1)(x+m_2).$
$\bf{\text{Solution Details:}}$
Factoring the $GCF=
13y
,$ the given expression, $
13y^3+39y^2-52y
,$ is equivalent to
\begin{array}{l}\require{cancel}
13y(y^2+3y-4)
.\end{array}
In the expression above, the value of $c$ is $
-4
$ and the value of $b$ is $
3
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-4\}, \{2,-2\},
\{-1,4\}, \{-2,2\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
1,-4
\}.$ Hence, the factored form of the given expression is
\begin{array}{l}\require{cancel}
13y(y+1)(y-4)
.\end{array}
