Answer
$(x-y+1)(x+y-1)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
x^2-y^2+2y-1
,$ group the last $3$ 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}
x^2-(y^2-2y+1)
.\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}
x^2-(y-1)^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}
[x-(y-1)][x+(y-1)]
\\\\=
[x-y+1][x+y-1]
\\\\=
(x-y+1)(x+y-1)
.\end{array}
