#### Answer

$(x+2m+n)(x-2m-n)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^2-4m^2-4mn-n^2
,$ group the last $3$ terms and factor the trinomial. Then use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the last $3$ terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
x^2-(4m^2+4mn+n^2)
.\end{array}
In the trinomial expression above, $a=
4
,b=
4
,\text{ and } c=
1
.$ Using the factoring of trinomials in the form $ax^2+bx+c,$ the two numbers whose product is $ac=
4(1)=4
$ and whose sum is $b$ are $\left\{
2,2
\right\}.$ Using these two numbers to decompose the middle term results to
\begin{array}{l}\require{cancel}
x^2-(4m^2+2mn+2mn+n^2)
\\\\=
x^2-[(4m^2+2mn)+(2mn+n^2)]
\\\\=
x^2-[2m(2m+n)+n(2m+n)]
\\\\=
x^2-[(2m+n)(2m+n)]
\\\\=
x^2-(2m+n)^2
.\end{array}
The expressions $
x^2
$ and $
(2mn+n)^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
x^2-(2m+n)^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}
(x)^2-(2m+n)^2
\\\\=
[x+(2m+n)][x-(2m+n)]
\\\\=
(x+2m+n)(x-2m-n)
.\end{array}