Factoring High School Level Olympiad Problem Factor $x^2 - 3xy + 2y^2 + x -8y - 6$
Attempt at a solution:
I have factored these and don't know how to continue...
$x^2-3xy +2y^2 = (x - y) (x-2y)$
$x^2 + x -6 = (x + 3) (x - 2)$
$2y^2 - 8y + 6 = 2 (y - 3)(y - 1)$ 
 A: Look for in form:
$$x^2-3xy+2y^2+x-8y-6=(x+Ay+B)(x+Cy+D)
$$
Plug $y=0$:
$$x^2+x-6=(x+B)(x+D) \Rightarrow B=3; D=-2.$$
Plug $x=0$:
$$2y^2-8y-6=(Ay+3)(Cy-2) \Rightarrow \begin{cases} AC=2 \\ -2A+3C=-8 \end{cases}$$
Can you finish?
Appendix: Note that the found parameters will not be suitable. So this method may not always work.
A: you can write your equation in the form $$y^2-y\left(4+\frac{3}{2}x\right)+\frac{x^2+x-6}{2}=0$$ and solve this for $y$
you will get
$$\left(y-\frac{3}{4}x-2-\frac{1}{4}\sqrt{x^2+40x+112}\right)\left(y+\frac{3}{4}x+2-\frac{1}{4}\sqrt{x^2+40x+112}\right)=0$$
A: Since the polynomial is of degree $2$, we can use the
well established "tool-set" for the study of Quadrics
or Conic Sections.
So
$$
\eqalign{
  & Q(x,y) = x^{\,2}  - 3xy + 2y^{\,2}  + x - 8y - 6 =   \cr 
  &  = \left( {x,y,1} \right)^T \left( {\matrix{
   1 & { - 3/2} & {1/2}  \cr 
   { - 3/2} & 2 & { - 4}  \cr 
   {1/2} & { - 4} & 6  \cr 
 } } \right)\left( {\matrix{
   x  \cr 
   y  \cr 
   1  \cr 
 } } \right) \cr} 
$$
But the determinant of the matrix defining the Conic
$$
{\bf A}_{\,Q}  = \left( {\matrix{
   1 & { - 3/2} & {1/2}  \cr 
   { - 3/2} & 2 & { - 4}  \cr 
   {1/2} & { - 4} & 6  \cr 
 } } \right)
$$
is not null, which means that the conic is not degenerate
and thus $Q(x,y)$ cannot be factored, not even in the complex field.
