Answer
$(2b+c-4)(2b+c+4)$
Work Step by Step
The first $3$ terms of the given expression, $
4b^2+4bc+c^2-16
,$ is a perfect square trinomial. Hence, the factored form is
\begin{array}{l}\require{cancel}
(4b^2+4bc+c^2)-16
\\\\=
(2b+c)^2-16
.\end{array}
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then the factored form of the expression, $
(2b+c)^2-16
,$ is
\begin{array}{l}\require{cancel}
[(2b+c)-4][(2b+c)+4]
\\\\=
(2b+c-4)(2b+c+4)
.\end{array}
