Show that $f(x)=x^2+x+4$ is irreducible over $\mathbb{Z}_{11}$ I know this can be done by evaluating $f$ at the points $0,1,...10$ to check if $f$ has a linear factor.
Is there any other shorter way?

We start with $$x^2+x+4=0$$ multiplying with $4$ we get $$4x^2+4x+16=0$$ This can be written as $$(2x+1)^2+15=0$$ modulo $11$ this is equivalent to $$(2x+1)^2-7=0$$
So we have to solve $$(2x+1)^2\equiv 7\mod 11$$
Using the quadratic reciprocity law we can calculate the legendre symbol
$$(\frac{7}{11})=-(\frac{11}{7})=-(\frac{4}{7})=-1$$
Hence $u^2\equiv 7\mod 11$ is not solveable because $7$ is not a quadratic residue modulo $11$. Since $x^2+x+4$ has no root in $\mathbb Z_{11}$, it follows that $x^2+x+4$ is irreducible in $\mathbb Z_{11}$
