#### Answer

$a(6-m)(5+m)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
30a+am-am^2
,$ factor first the $GCF.$ Then to factor the resulting trinomial, 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 $GCF=a,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
a(30+m-m^2)
.\end{array}
To factor the trinomial expression above, note that the value of $ac$ is $
-1(30)=-30
$ and the value of $b$ is $
1
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-30\}, \{2,-15\}, \{3,-10\}, \{5,-6\},
\\
\{-1,30\}, \{-2,15\}, \{-3,10\}, \{-5,6\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
-5,6
\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
a(30-5m+6m-m^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}
a[(30-5m)+(6m-m^2)]
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
a[5(6-m)+m(6-m)]
.\end{array}
Factoring the $GCF=
(6-m)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
a[(6-m)(5+m)]
\\\\=
a(6-m)(5+m)
.\end{array}