Factoring multivariate polynomial I'm trying to factor
$$x^3+x^2y-x^2+2xy+y^2-2x-2y \in \mathbb{Q}[x,y].$$
The hint for the exercise is to use the recursive multivariate polynomial form. So I'm using $\mathbb{Q}[x][y]$:
$$ x^3 + x^2(y-1) + x^1(y-2) + x^0(y^2-2y) $$
At this point I am stuck. Are their any general techniques to do this by hand?
 A: Let’s go back to the original polynomial,
$$x^3+x^2y-x^2+2xy+y^2-2x-2y\;.\tag{1}$$
That $2xy$ looks a lot like the middle term of $(x+y)^2$, and the $-2x-2y$ can certainly be written nicely in terms of $x+y$, so let’s try something along those lines. $(x+y)^2=x^2+2xy+y^2$, and we have the $2xy+y^2$, but instead of $x^2$, we have $-x^2$. In other words, we have $(x+y)^2-2x^2$. That’s not entirely promising, but let’s see where it goes. We can rewrite $(1)$ as
$$\begin{align*}
(x+y)^2&-\;2x^2-2(x+y)+x^3+x^2y\\\\
&=(x+y)^2-2(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x+y)+x^2(x+y-2)\\\\
&=(x+y-2)(x^2+x+y)\;.
\end{align*}$$
Added: Alternatively, following the hint, interchange the rôles of $x$ and $y$, rewriting $(1)$ as $$y^2+(x^2+2x-2)y+(x^3-x^2-2x)\;.\tag{2}$$
Remember that linear factors of a polynomial correspond to zeroes. Set $(2)$ equal to $0$ and use the quadratic formula to solve for $y$: 
$$\begin{align*}
&(x^2+2x-2)^2-4(x^3-x^2-2x)\\
&\qquad=x^4+4x^3-8x+4-4x^3+4x^2+8x\\
&\qquad=x^4+4x^2+4\\
&\qquad=(x^2+2)^2\;,
\end{align*}$$
so $$y=\frac{-x^2-2x+2\pm(x^2+2)}2\;,$$
and $y=-x^2-x$ or $y=-x+2$. Thus, $(2)$ factors as $$\Big(y-(-x^2-x)\Big)\Big(y-(-x+2)\Big)=(y+x^2+x)(y+x-2)\;.$$
A: A detailed explanation can be found here: Factorize the polynomial $x^3+y^3+z^3-3xyz$
The calculations for this example:
$$x^3+x^2y-x^2+2xy+y^2-2x-2y$$ can be viewed as is a polynomial of $\mathbb{Q}[y][x]$
so $$p(x)=x^3+(y-1)x^2+(2y-2)x+(y^2-2y)$$
If this monic polynomial in $x$ of degree 3 is not irreducible it has a linear factor and therefore it has a zero.
This zero must be a factor of the constant term $y(y-2)$. Checking all possibel values one finds $p(-y+2)=0$. So $x+y-2$ is a factor. Dividing $p(x)$ by $x+y-2$ gives $y+x^2+x$ which is irreducible.
