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...
 A: Consider the numerator $2x^2-(y+z)(y+z-x).$ The first thing to notice here is the sum $y+z.$ You may set $y+z=p$ if it makes the following more convenient to follow. Now expanding gives $$2x^2-(y+z)^2+x(y+z).$$ You almost have a difference of squares -- indeed you do already, but it doesn't help much, so we split into four terms so we can group in pairs, as follows $$x^2-(y+z)^2+x^2+x(y+z)=(x-y-z)(x+y+z)+x(x+y+z)=(x+y+z)(x-y-z+x),$$ or $(x+y+z)(2x-y-z).$
A similar procedure, or just performing the transformations $x\mapsto y$ and $y\mapsto x,$ gives the denominator in factored form as $(x+y+z)(2y-x-z).$
A: $\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)(x+y+z)}{(2y-x-z)(x+y+z)}=\frac{(2x-y-z)}{(2y-x-z)}.
$
A: 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.
