What is the easiest way to factorize $x^2-xy-2y^2-x-4y-2=0$ without using any software?

Question

I already tried to create a small group between any variables like x and y to make it distributiveable. But nothing happened, and I cannot continue to the desired form.

Here's my attempt: I think I need to factorize $x^2-xy-2y^2$ such that I get $(x-2y)(x+y)$ , but then I got the bad form like $(x-2y)(x+y)-(x+4y+2)$ so that I cannot continue to find a better form to perform distributive property here.

Is there any theorems or any ideas to factorize such that equation like this?

Thanks

Answer 1

A genral method for the quadratic of $x,y$:

Take it as a quadratic of $x$ and treat $y$ as constant, then $$x^2-x(-y-1)-2y^2-4y-2=0$$ $$ \implies x=\frac{(y+1)\pm\sqrt{(y+1)^2+4(2y^2+4y+2)}}{2}$$ $$x=\frac{(y+1)\pm 3(y+1)}{2}$$ You get two linear factors here.

Answer 2

As also suggested in the comments, we can proceed as follows

$$(ax+by+c)(dx+ey+f)=adx^2+bey^2+(ae+bd)xy+(af+cd)x+(bf+ce)y+cf$$

and then

• $ad=1$
• $be=-2$
• $ae+bd=-1$
• $af+cd=-1$
• $bf+ce=-4$
• $cf=-2$

and we can try to solve starting with $a=d=1$ or $a=d=-1$.

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As an alternative by $y=ax+b$ we obtain

$$x^2-xy-2y^2-x-4y-2=$$

$$=x^2-ax^2-bx-2a^2x^2-4abx-2b^2-x-4ax-4b-2=$$

$$=(1-a-2a^2)x^2+(-b-4ab-1-4a)x+(-2b^2-4b-2)=0$$

which requires

• $1-a-2a^2=0 \implies a=-1 \quad a=\frac12$
• $-2b^2-4b-2=0 \implies b=-1$
• $-b-4ab-1-4a=0$ always true for $b=-1$

therefore two factors are

• $c(y+x+1)$
• $d(2y-x+2)$

with $c$ and $d$ constants easy to find.