Answer
$(5c-2+d)(5c-2-d)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
25c^2-20c+4-d^2
,$ group the first $3$ terms since these 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.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms above results to
\begin{array}{l}\require{cancel}
(25c^2-20c+4)-d^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}
(5c-2)^2-d^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}
[(5c-2)+d][(5c-2)-d]
\\\\=
(5c-2+d)(5c-2-d)
.\end{array}