irreducibility in $\mathbb Q[X]$ Are these polynomials irreducible in $\mathbb Q[X]$:
1) $x^4+3x^3+x^2-2x+1$
What I did: I reduced it modulo $3$, then saw that it has no roots, so then I checked all $9$ monic polynomials of $\mathbb F_3$ and got $3$ of them, then checked their values under multiplication (as I calculated them allready to determin if they are reducible) with each other to conclude that they have different values then 1) for atleast one $a\in\mathbb F_3$. This way seems very inefficient, is it possible to do this more clever?
2) $2x^4+200x^3+2000x^2+20000x+20$
Here I divided by $2$ and then I could use Schönemann's criterion.
3) $x^2y+xy^2-x-y+1$ in $\mathbb Q[X,Y]$
$x y^2+(x+1)(x-1)y+(-x+1)$, so after Schönemann's criterion it's irreducible.
 A: (3) To "see" it better I would change $x=x+y$ and $y=x-y$ to get 
$$P(x,y)=2x^3-2yx^2-2(y^2-2)x+2y^3+1$$
So, a non-trivial factorization would have to be of the form $2(x+p(y))(x^2+q(y)x+r(y))$. 
From this $p(y)r(y)=2y^3+1$. But $2y^3+1$ is irreducible over $\mathbb{Q}$. So, either $p(y)$ or $r(y)$ is a constant.
If $p(y)$ where a constant $c$, then $P(c,y)=0$ but this can't be since $P(c,y)$ has degree $3$ in $y$.
Therefore we would need $p(y)=-a(2y^3+1)$ and $r(y)=-a^{-1}$ for some non-zero $a$. But when we try $$0\equiv P(a(2y^3+1),y)=16a^3y^9+24a^3y^6-8 a^2 y^7-4ay^5-8a^2y^4+(12a^3-4a+2)y^3-2ay^2-2a^2y-2a+1+2 a^3$$
We see that $a$ must be zero.

(1) It doesn't have rational roots since the only candidates $\pm1$ do not work. Therefore a factorization would have to be of the form
$$\begin{align}x^4+3x^3+x^2−2x+1&=(x^2+ax+b)(x^2+px+q)\\&=x^4+(a+p)x^3+(ap+b+q)x^2+(aq+bp)x+bq\end{align}$$
Therefore either $b=q=1$ or $b=q=-1$.
Now we can use the equations (coming from equating coefficients above) $$\begin{align}a+p&=3\\ap&=1-(b+q)=-1\text{ or }3\end{align}$$
to find what should be the values of $a,p$.
They should be roots of the polynomial $$x^2-3x+(-1\text{ or }3)$$ but none of these has rational roots.
