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