Answer
$6mn^2(m-5)^2$
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 $GCF=
6mn^2
,$ the given expression, $
6m^3n^2-60m^2n^3+150mn^4
,$ is equivalent to
\begin{array}{l}\require{cancel}
6mn^2(m^2-10mn+25n^2)
.\end{array}
In the expression above, the value of $c$ is $
25
$ and the value of $b$ is $
-10
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,25\}, \{5,5\},
\{-1,-25\}, \{-5,-5\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-5,-5
\}.$ Hence, the factored form of the given expression is
\begin{array}{l}\require{cancel}
6mn^2(m-5)(m-5)
\\\\=
6mn^2(m-5)^2
.\end{array}
