Factor this quadratic expression

Question

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?

Answer 1

You almost got it, but you made a couple of careless errors. You should have put $0=Ax^2-Dyx-Ay^2$, and then you forgot that the “constant” term at the right is not $-A$ but $-Ay^2$. So you should have gotten $$ x=\frac{Dy\pm\sqrt{D^2y^2+4A^2y^2}}{2A}=y\frac{D\pm\sqrt{D^2+4A^2}}{2A}\,. $$ Let’s call the two possible constants that multiply $y$ on the right $K_1$ and $K_2$. then the factorizatation is $(x-K_1y)(x-K_2y)$, except that you need a correcting constant in front, I guess it’s $A$.

Answer 2

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 $$\frac{1}{4A}\left(2Ax-Ey -\sqrt{4A^2+E^2}\,y\right)\left(2Ax-Ey +\sqrt{4A^2+E^2}\,y\right).$$