How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer. 
It says that 
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate on how that is possible? What am I not seeing...

From the numerator, you can pick up a $2x-y-z$, obtaining:
$$ x^2+x y-y^2+x z-2 y z-z^2=(2x-y-z)\cdot(x+y+z)$$
From the denominator, you can pick up $2y-x-z$, arriving at:
$$2 y^2-x^2+x y-2 x z+y z-z^2=(2y-x-z)\cdot(x+y+z)$$
Now, you can rewrite your fraction as:
$$\frac{(2 x^2+x y-y^2+x z-2 y z-z^2)}{(2 y^2-x^2+x y-2 x z+y z-z^2)}=\frac{(2x-y-z)\cdot(x+y+z)}{(2y-x-z)\cdot(x+y+z)}$$
If $x+y+z\neq 0$, you can cancel out the common factor $x+y+z$ and arrive at:
$$\frac{(2x-y-z)}{(2y-x-z)}$$
This fraction can't further be simplified.
