#### Answer

$(2r+s+t)(2r-s-t)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
4r^2-s^2-2st-t^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}
4r^2-(s^2+2st+t^2)
.\end{array}
In the trinomial expression above, $b=
2
,\text{ and } c=
1
.$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{
1,1
\right\}.$ Using these two numbers, the factored form of the expression above is
\begin{array}{l}\require{cancel}
4r^2-[(s+1t)(s+1t))
\\\\=
4r^2-(s+t)^2
.\end{array}
The expressions $
4r^2
$ and $
(s+t)^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
4r^2-(s+t)^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}
(2r)^2-(s+t)^2
\\\\=
[2r+(s+t)][2r-(s+t)]
\\\\=
(2r+s+t)(2r-s-t)
.\end{array}