Factoring $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6$

Question

This is quite easy to do if we do the middle term splitting method, but here's another method given in the book:

Only considering terms with powers of $x$ and constants:
$-3x^2 + 11x - 6 = (3x - 2)(-x + 3)$

Only considering terms with powers of y and constants:
$8y^2 - 8y - 6 = (-4y - 2)(-2y + 3)$

Merging both we get $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6 = (3x - 4y - 2)(-x - 2y + 3)$

Why does this method work, and is there a name to this method?

Answer 1

So you want: $$-3x^2 - 2xy + 8y^2 + 11x - 8y - 6= (ax+by+c)(dx+ey+f)$$

If we set $y=0$ we get $$\boxed{-3x^2+11x-6 = (ax+c)(dx+f)}$$

and since $$-3x^2+11x-6 = (-3x+2)(x-3)$$ so $a=-3$, $c=2$, $d=1$ and $f=-3$ (or $a=3$, $c=-2$, $d=-1$ and $f=3$.)

Seting $x= 0$ we get $$\boxed{ 8y^2-8y-6 = (by+c)(ey+f)}$$ so ...