It's a part of a bigger problem I'm facing. Not only I don't know how to prove it, I don't know if it's true or false at all (so I have no idea what to try to prove and so I don't know where to start).
My biggest problem is on the field, I feel like that having only the characteristic as information isn't enough to give an answer.
PS: sorry for my bad English.
PPS: I don't know if it's wrong, but I tagged it also with Galois-theory because I met this polynomial studying it.
On
Hint: Let $K$ be a field of characteristic $p$, then $\mathbb{F}_p\subseteq K$. Since the Frobenius map ($x\mapsto x^p$) is constant on $\mathbb{F}_p$, we know that for $x\in\mathbb{F}_p$, $x^p=x$.
Therefore, a root of $x^p-ax-b$ is the same as a root of $(1-a)x-b$. Hence, if $a\not=1$, then one can find a root of the given expression.
If it is irreducible over $\mathbb F_q$, then (at least) it has no root in $\mathbb F_q$. (where $q$ is a $p$-power).
Define $b := 1-a$ (where $a\ne 0,1$). Then $$f(1)=1-a-(1-a)=0$$ which implies $$(x-1) \mid (x^p-ax-b).$$ We also need to mention that degree $p>1$ for having more factors. But every prime is bigger than $1$.
If $p=2$, $a=1$ and $b=1$, then $$x^2+x+1$$ is irreducible over $\mathbb F_{2^n}$ where $n$ is odd and reducible over $\mathbb F_{2^n}$ where $n$ is even.