#### Answer

$(x+1+2z)(x+1-2z)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^2+2x+1-4z^2
,$ group the first $3$ terms since these form a perfect square trinomial. Then, factor the trinomial. This results to an expression which is a difference of $2$ squares. Finally, use the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first $3$ terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x^2+2x+1)-4z^2
.\end{array}
Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $
1(1)=1
$ and the value of $b$ is $
2
.$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{
1,1
\right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to
\begin{array}{l}\require{cancel}
(x^2+x+x+1)-4z^2
.\end{array}
Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
[(x^2+x)+(x+1)]-4z^2
.\end{array}
Factoring the $GCF$ in each group results to
\begin{array}{l}\require{cancel}
[x(x+1)+(x+1)]-4z^2
.\end{array}
Factoring the $GCF=
(x+1)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
[(x+1)(x+1)]-4z^2
\\\\=
(x+1)^2-4z^2
.\end{array}
The expressions $
(x+1)^2
$ and $
4z^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
(x+1)^2-4z^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+1)^2-(2z)^2
\\\\=
[(x+1)+2z][(x+1)-2z]
\\\\=
(x+1+2z)(x+1-2z)
.\end{array}