How to factorise this expression $ x^2-y^2-x+y$

Question

This part can be factorised as $x^2-y^2=(x+y)(x-y)$, How would the rest of the expression be factorised ?

:)

Answer 1

$$x^2-y^2-x+y=(x^2-y^2)-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1)$$

Answer 2

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 $x$.

$$x^2-x+(y-y^2)=0$$ $$x=\frac{1\pm\sqrt{1+4(y^2-y)}}{2}$$ $$\left(x-\frac{1+\sqrt{1+4(y^2-y)}}{2}\right)\left(x-\frac{1-\sqrt{1+4(y^2-y)}}{2}\right)$$ $$\frac{1}{4}\left(2x-1+\sqrt{(2y-1)^2}\right)\left(2x-1-\sqrt{(2y-1)^2}\right)$$ $$\left(x+y-1\right)\left(x-y\right)$$

Answer 3

If you have already $x^2-y^2=(x+y)(x-y)$, then $$ x^2-y^2-(x-y)=(x-y)(x+y)-(x-y)=(x-y)(x+y-1). $$

Answer 4

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

$$ x^2 + y^2 - x +y = (x+y)(x-y)-(x-y) = (x+y-1)(x-y) $$

Answer 5

Another method to factor $${ x }^{ 2 }-x+\frac { 1 }{ 4 } -{ y }^{ 2 }+y-\frac { 1 }{ 4 } ={ \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( y-\frac { 1 }{ 2 } \right) }^{ 2 }=\left( x-\frac { 1 }{ 2 } +y-\frac { 1 }{ 2 } \right) \left( x-\frac { 1 }{ 2 } -y+\frac { 1 }{ 2 } \right) =\left( x-y \right) \left( x+y-1 \right) $$

Answer 6

Notice here,

$$x^2 - y^2 - x + y$$ $$(x+y)(x-y)-1(x-y)$$ $$(x-y) (x+y-1)$$ That's done.