Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$ I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as
$$p(x,y)=(y+a)(y^2+by+c) = y^3 + (a+b)y^2 + (ab+c)y + ac$$
for some $a,b,c \in \mathbb{Z}[x]$.
Comparing coefficients we have 
\begin{align}
&a+b=x \\
&ab+c = x^2+x \\
&ac = 4x^3+2x = 2x(2x^2+1)
\end{align}
From the last equation we have $a = 2x$ and $c=2x^2+1$ or alternatively $a = 2x^2+1$ and $c=2x$. 
If $a=2x$, $c=2x^2+1$ then from the first equation, $2x+b=x$ so $b=-x$. From the second equation, $-2x^2+x-1=x^2+x$, but this doesn't hold for all $x$... 
If $a=2x^2+1$, $c=2x$ then from the first equation, $2x^2+1+b=x$ so $b=-2x^2+x-1$. From the second equation, $(2x^2+1)(-2x^2+x-1) = x^2+x$, but again the l.h.s. and r.h.s. don't match since without even computing we see the l.h.s. is degree $4$.
Is there a better way to prove this?

Rewrite: 
$$p(x,y)= y^3+xy^2+x(x+1)y+2x(2x^2+1), $$
and as $p(x,y)\in \mathbf Z[x][y]$, we can use the generalisation of Eisenstein's criterion directly:
$(x)$ is a prime ideal in the U.F.D. $\mathbf Z[x]$, and


*

*$x$ divides divides all coefficients, but the leading coefficient;

*$x^2$ does not divide the constant term.
Hence $p(x,y)$ is irreducible in $\mathbf Z[x][y]\simeq\mathbf Z[x,y]$.

