#### Answer

$(m+n-5)(m-n+1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Group the first $3$ terms and the last $3$ terms of the given expression, $
m^2-4m+4-n^2+6n-9
,$ and use the factoring of perfect square trinomials. Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms and last $3$ terms of the given expression results to
\begin{array}{l}\require{cancel}
(m^2-4m+4)-(n^2-6n+9)
.\end{array}
Using the factoring of perfect square trinomials which is given by $a^2-2ab+b^2=(a-b)^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(m-2)^2-(n-3)^2
.\end{array}
The expressions $
(m-2)^2
$ and $
(n-3)^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
(m-2)^2-(n-3)^2
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(m-2)+(n-3)][(m-2)-(n-3)]
\\\\=
[m-2+n-3][(m-2-n+3]
\\\\
[m+n-5][m-n+1]
\\\\=
(m+n-5)(m-n+1)
.\end{array}