Irreducibility of $x^2+x+4\in {\Bbb Z}_p[x]$ over ${\Bbb Z}_p$? The following is an exercise in abstract algebra.

Show that $x^2+x+4$ is irreducible over ${\Bbb Z}_{11}$.

One test all the elements in ${\Bbb Z}_{11}$ to show that $x^2+x+4$ has no zeros in ${\Bbb Z}_{11}$.
Here is my question:

Can one know the irreducibility of $x^2+x+4\in{\Bbb Z}_p[x]$ for the general field ${\Bbb Z}_p$ where $p$ is prime?

 A: Yes. More exactly, this Polynomial is sometimes reducible and sometimes irreducible, and you can easily find a condition: $f(x)$ irreducible if and only if $p$ is of a certain form.
$$\Delta=1-16=-15 \,.$$
As Zev explained in his answer, in the case $p \neq 2$ you can find the quadratic formula by completing the square, while in the case $p=2$ it is trivial to check that our polynomial is reducible.
It follows that $x^2+x+4$ is irreducible in $\mathbb Z_{p}$ if and only if $\left(\frac{-15}{p}\right)=-1$ where this is the Legendre symbol.
Now, since
$$\left(\frac{-15}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{3}{p}\right)\left(\frac{5}{p}\right)$$
using quadratic residues you can easily find all primes for which this is $-1$.
Added: You need to leave out the cases $p=2, p=3, p=5$, it is easy to check what happens then. otherwise
$$\left(\frac{-1}{p}\right)=(-1)^\frac{p-1}{2}$$
$$\left(\frac{3}{p}\right)=1 \, \mbox{iff} \, p=\pm 1 \pmod{12}$$
$$\left(\frac{3}{p}\right)=-1 \, \mbox{iff} \, p=\pm 5 \pmod{12}$$
$$\left(\frac{5}{p}\right)=1 \, \mbox{iff} \, p=\pm 1 \pmod{5}$$
$$\left(\frac{5}{p}\right)=-1 \, \mbox{iff} \, p=\pm 2 \pmod{5}$$
So, now all you have to do is check the cases $p=2,3,5$ and then study $p \pmod{60}$.
A: Hint $\rm\ \ f(x) = x^2 + x + 4\ $ is irreducible over $\rm\,\Bbb Z/p\,$ iff its discriminant $-15$ is not a square $\rm\,mod\ p.\:$ By quadratic reciprocity, $\rm\,(-15|p) = (-3|p)(5|p),\:$ and $\rm\,(-3|p) = 1\iff p\equiv 1\,\ (mod\ 6),\:$ and $\rm\,(5|p) = 1\iff p\equiv \pm1\,\ (mod\ 5).\:$ Thus $\rm\,(-3|p)(5|p) = -1\,$ if and only if 
$$\begin{eqnarray} \rm\,p\equiv +1\,\ (mod\ 6),\ p\equiv \pm2\,\ (mod\ 5) &\iff& \rm\,p\equiv \ \,7,\ 13\,\ (mod\ 30)\\
\rm\,p\equiv -1\,\ (mod\ 6),\ p\equiv \pm1\,\ (mod\ 5) &\iff& \rm\,p\equiv 11,-1\,\ (mod\ 30)
\end{eqnarray}$$
Therefore $\rm\ x^2 + x + 4\:$ is irreducible $\rm\,mod\ p\ \iff\ p\equiv 7,11,13,29\,\ (mod\ 30).$
