How to factorise this expression $ x^2-y^2-x+y$ This part can be factorised as  $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?
:)
 A: $$x^2-y^2-x+y=(x^2-y^2)-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1)$$
A: If you have already $x^2-y^2=(x+y)(x-y)$, then
$$
x^2-y^2-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1).
$$
A: Another method to factor $${ x }^{ 2 }-x+\frac { 1 }{ 4 } -{ y }^{ 2 }+y-\frac { 1 }{ 4 } ={ \left( x-\frac { 1 }{ 2 }  \right)  }^{ 2 }-{ \left( y-\frac { 1 }{ 2 }  \right)  }^{ 2 }=\left( x-\frac { 1 }{ 2 } +y-\frac { 1 }{ 2 }  \right) \left( x-\frac { 1 }{ 2 } -y+\frac { 1 }{ 2 }  \right) =\left( x-y \right) \left( x+y-1 \right) $$
A: This is a far less elegant method than the other answers but I'll keep it up for posterity: set it up as a quadratic in $x$.
$$x^2-x+(y-y^2)=0$$
$$x=\frac{1\pm\sqrt{1+4(y^2-y)}}{2}$$
$$\left(x-\frac{1+\sqrt{1+4(y^2-y)}}{2}\right)\left(x-\frac{1-\sqrt{1+4(y^2-y)}}{2}\right)$$
$$\frac{1}{4}\left(2x-1+\sqrt{(2y-1)^2}\right)\left(2x-1-\sqrt{(2y-1)^2}\right)$$
$$\left(x+y-1\right)\left(x-y\right)$$
A: Notice here,
$$x^2 - y^2 - x + y$$
$$(x+y)(x-y)-1(x-y)$$
$$(x-y) (x+y-1)$$
That's done.
A: You can observe that $-x+y = -(x-y)$. So both $x^2-y^2$ and $-x+y$ have $x-y$ as a factor. You can use these info to get below factorization
$$
x^2 + y^2 - x +y = (x+y)(x-y)-(x-y) = (x+y-1)(x-y)
$$
