Answer
$(2+k-h)(2-k+h)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
-k^2-h^2+2kh+4
,$ rewrite and group terms that will form a perfect square trinomial. Then factor the trinomial. The resulting expression becomes a difference of $2$ squares. Factor this expression using the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
The given expression can be rewritten as
\begin{array}{l}\require{cancel}
4-k^2+2kh-h^2
.\end{array}
Grouping the last $3$ terms above results to
\begin{array}{l}\require{cancel}
4-(k^2-2kh+h^2)
.\end{array}
The trinomial above is a perfect square trinomial. Using $a^2\pm2ab+b^2=(a\pm b)^2,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
4-(k-h)^2
.\end{array}
The expression above is a difference of $2$ squares. Using $a^2-b^2=(a+b)(a-b),$ the factored form of the expression above is
\begin{array}{l}\require{cancel}
[2+(k-h)][2-(k-h)]
\\\\=
[2+k-h][2-k+h]
\\\\=
(2+k-h)(2-k+h)
.\end{array}