Showing that $x^4 + 4x^3 + 11x^2 + 10x + 9$ is irreducible over $\mathbb{Q}$ 
Let $p(x) = x^4 + 4x^3 + 11x^2 + 10x + 9$. How to how that it is irreducible over $\mathbb{Q}$?

The first thing I tried was Eisenstein, but that obviously doesn't work with the polynomial in that form. 
Then I looked at $p$ in $\mathbb{F}_2$ and $\mathbb{F}_5$ in order to show that it irreducible in one of them and then to use Gauß' Theorem. However, it turns out that $p$ is reducible over both fields.
I've seen people use nifty substitutions in connection with the Eisenstein criterion. I tried to represent $p$ in terms of $x+1$, but that doesn't seem to work, either.
Now I'm at my wit's end. What else could I try? 
 A: It is a standard result that it is irreducible over Q iff it is irreducible over Z. If it factors as linear x cubic, then the linear factor would have to be $x+a$ for some integer $a$, so $a$ would be a root and a factor of 9. So you have to check 1, 3, -3, 9, -9. None of them are roots. If it factors as quadratic x quadratic, then it is $(x^2+ax+b)(x^2+cx+d)$.
So you have $a+c=4,bd=9,b+d+ac=11$. If $b=1,-1,9$ or -9, then $ac=1$ or 21, both incompatible with $a+c=4$. So $b=d=3$ or $b=d=-3$, so $ac=5$ or 17. Neither is compatible with $a+c=4$. So it is irreducible.
A: If in addition to knowing that $p$ is reducible both modulo $2$ and modulo $5$ we have the factorizations of $p$ modulo those primes into irreducible polynomials, namely,
$$p(x) \equiv (x^2 + x + 1)^2 \pmod 2 \qquad \textrm{and} \qquad p(x) 
\equiv (x^3 + x + 1)(x + 4) \pmod 5,$$
we see immediately that if $p$ were reducible it would have both an irreducible quadratic factor and an irreducible cubic factor, which (because $\deg p = 4$) is impossible, hence $p$ is irreducible over $\Bbb Q$. (As Aryaman Maithani points out in the comments, $p(x)$ also factors modulo $3$ into the product of a cubic factor and a linear one, and factoring modulo $3$ is typically faster that factoring modulo $5$, leading to a probably faster solution using the same reasoning.)
