Answer
$(6m-5)(-m-3)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping.
$\bf{\text{Solution Details:}}$
In the given expression, $
-6m^2-13m+15
,$ the value of $ac$ is $
-6(15)=-90
$ and the value of $b$ is $
-13
.$
The possible pairs of integers whose product is $ac$ are
\begin{array}{l}\require{cancel}
\{1,-90\}, \{2,-45\}, \{3,-30\}, \{5,-18\}, \{6,-15\}, \{9,-10\},
\{-1,90\}, \{-2,45\}, \{-3,30\}, \{-5,18\}, \{-6,15\}, \{-9,10\}
.\end{array}
Among these pairs, the one that gives a sum of $b$ is $\{
5,-18
\}.$ Using these $2$ numbers to decompose the middle term of the given expression results to
\begin{array}{l}\require{cancel}
-6m^2+5m-18m+15
.\end{array}
Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(-6m^2+5m)-(18m-15)
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
-m(6m-5)-3(6m-5)
.\end{array}
Factoring the $GCF=
(6m-5)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(6m-5)(-m-3)
.\end{array}
