<s>[INST] Factor this quadratic expression I need to do the following:

Prove that a quadratic expression of the form $A(x^2-y^2) - (B-C)xy$ can be always factored into two linear factors.

It is easy enough to compare this with the standard representation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c$ and say that the condition $abc + 2fgh -af^2 - bg^2 - ch^2 = 0$ is satisfied. 
However, what are those two linear factors?
At best the given expression can be written as $Ax^2 - Ay^2 - Bxy + Cxy$ or $x(Ax-By) -y(Ax-Cy)$. But this doesn't look like it can be always factored.
Second try:
Let $B-C = D$ for convenience. 
Now the expression is:
$Ax^2 - Ay^2 - Dxy$
$= Ax^2 - (Dy)x + Ay^2$
If this be equated to zero,
$x = \dfrac{Dy \pm \sqrt{(Dy)^2 - 4A^2}}{2A}$
$\Rightarrow 2Ax - Dy = \pm \sqrt{(Dy)^2 - 4A^2}$
Squaring both sides and cancelling $D^2y^2$, we are left with
$Ax^2 - Dxy + A = 0$
Sigh! Looks like I'm just going in circles, and while it's easy to prove using the standard results, I can't seem to find the factors. How can this be done?
 [/INST] You want to factor $A(x^2-y^2) - (B-C)xy$. To ease typing, write $E$ for $B-C$.
Rewrite our quadratic as 
$$\frac{1}{4A}\left(4A^2 x^2-4AExy -4A^2y^2\right).$$
Complete the square. We get
$$\frac{1}{4A}\left((2Ax-Ey)^2  -(4A^2+E^2)y^2\right).$$
This is a difference of squares, and we get
$$\frac{1}{4A}\left(2Ax-Ey -\sqrt{4A^2+E^2}\,y\right)\left(2Ax-Ey +\sqrt{4A^2+E^2}\,y\right).$$
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