Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will 
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
 A: There is no simple factorization of $x^2+2xy-y^2$ nor $x^2-2xy-y^2$, although you can write:
$$\begin{align}x^2+2xy-y^2 &= (x+y)^2-2y^2 \\&= \left(x+y(1+\sqrt{2})\right)\left(x+y(1-\sqrt{2})\right)
\end{align}$$
and similarly:
$$x^2-2xy-y^2 = \left(x-y(1+\sqrt{2})\right)\left(x-y(1-\sqrt{2})\right)$$
A: You can use the quadratic formula:
$$x^2 + (-2y)x + (-y^2) = 0\to \\ x= \frac{2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$
So $x^2 - 2xy - y^2 = [x - (1+\sqrt{2})y][x - (1-\sqrt{2})y].$
For the other case,
$$x^2 + (2y)x + (-y^2) = 0\to \\ x= \frac{-2y \pm \sqrt{4y^2 - 4(1)(-y^2)}}{2} \\ =-y \pm\sqrt{2}y = (1 \pm \sqrt{2})y.$$
So $x^2 + 2xy - y^2 = [x - (1-\sqrt{2})y][x - (1+\sqrt{2})y].$
A: This is another way of solving this problem using Completing the Square method.
$x^2+2xy-y^2=??$
First we have,
$x^2+2xy-y^2=0$
$x^2+2xy=y^2,$
By adding both sides by $y^2$ to make it perfect square then,
$x^2+2xy+y^2=y^2+y^2$
$(x+y)^2=2y^2$
$x+y=[2^(1/2)]y$
$x=(+ or -)[2^(1/2)]y-y$
$x=[2^(1/2)]y-y  ; x=-[2^(1/2)]y-y $
$x=[2^(1/2)-1]y   ; x=-[2^(1/2)-1]y$
therefore we have,
${x-[2^(1/2)-1]y}{x+[2^(1/2)-1]y}$
By solving 
$x^2-2xy-y^2=??$
Try to solve this by relying with this pattern. Thank you.
A: You can factor a quadratic trinomial
$$ax^2+bxy+cy^2$$
by finding the roots of
$$\frac{ax^2+bxy+cy^2}{y^2}=a\left(\frac xy\right)^2+b\left(\frac xy\right)+c=0.$$
Then
$$ax^2+bx+c=y^2a\left(\frac xy-r_0\right)\left(\frac xy-r_1\right)=a\left(x-r_0y\right)\left(x-r_1y\right).$$
