I am a completing a past paper question and I am undecided on what method to use here. The question is:
For what $a$ is $f(x)=x^3+x+a\in\mathbb{Z}_{7}[x]$ irreducible? My ideas are:
(1) Check each $a\in\{0,1,...6\}$ individually. I tried this but was not sure how I would prove reducibility individually, if it were the case.
(2) Perhaps there is some result regarding reducibility in $\mathbb{Z}_{p}[x]$. I am not sure if this is the case.
Any suggestions would be much appreciated.
Here is a table of the values of $x\to x^3+x$ for $x$ in the field with seven elements $\Bbb F_7$: $$ \begin{array}{r||c|c|c|c||c|c|c} x & 0 & 1 & 2 & 3 & -3 & -2 & -1 \\\hline x^3+x & 0 & 2 & 3 & 2 & -2 & -3 & -2 \end{array} $$ So for $a$ among $0$, $\pm 2$, $\pm 3$ the polynomial $x^3+x+a$ has at least one root in $\Bbb F_7$, so it is not irreducible. Else, the polynomial has no root in $\Bbb F_7$, thus is irreducible: $$ \begin{aligned} x^3 + x + 0 &= x (x^2 + 1) &&\text{ reducible,}\\ x^3 + x + 1 &= x^3 + x + 1 &&\text{ irreducible,}\\ x^3 + x + 2 &= (x + 1) (x + 3)^2 &&\text{ reducible,}\\ x^3 + x + 3 &= (x + 2) (x^2 + 5x + 5) &&\text{ reducible,}\\ x^3 + x + 4 &= (x + 5) (x^2 + 2x + 5) &&\text{ reducible,}\\ x^3 + x + 5 &= (x + 6) (x + 4)^2 &&\text{ reducible,}\\ x^3 + x + 6 &= x^3 + x + 6 &&\text{ irreducible.} \end{aligned} $$