The complete problem appears in Hungerford's Algebra.
Let $c\in F$, where $F$ is a field of characteristic $p$ ($p$ prime). Then $x^p−x−c$ is irreducible in $F[x]$ if and only if $x^p−x−c$ has no root in $F$.
One direction has already been answered:
$x^p -x-c$ is irreducible over a field of characteristic $p$ if it has no root in the field
I wish to prove the other direction:
If $x^p−x−c$ is irreducible in $F[x]$, then $x^p−x−c$ has no root in $F$.
I tried to prove this using the contrapositive:
Assume $\alpha\in F$ such that $\alpha^p−\alpha−c=0$. We show that we can factor $x^p−x−c$ into simpler polynomials, $f,g$ each of degree $<p$. I am stumped because I cannot show such a polynomial factorization.
I was thinking that maybe I could say that, hence, $x-\alpha$ is a linear factor of $x^p−x−c$, and thus, $x^p−x−c = (x-\alpha)f(x)$ where $\deg f(x) \le p-1$.
Therefore, $x^p−x−c$ is a reducible polynomial.
When $\mathrm{char} F = 0$, then this statement is supposedly false. However, I cannot think of a reducible polynomial over $\mathbb Q$ of degree $5$ whose roots are not rational, or an irreducible polynomial over $\mathbb Q$ of degree $5$ but has roots in $\mathbb Q$.
Best Answer
The contrapositive of your statement is: If $f(x)$ has a root in $F$, then $f(x)$ is reducible over $F$. This is true over any field.
Over any field you have a factorization theorem: If $f\in F[x]$ and $\alpha\in F$, then $f(x)=q(x)(x-\alpha)+f(\alpha)$. In particular, if $\alpha$ is a root of $f$, then $(x-\alpha)$ is a factor of $f$.
The way this is proved is to compute $f(x)-f(\alpha)$ and use the factorization $$x^k-\alpha^k=(x-\alpha)(x^{k-1}+x^{k-2}\alpha+\cdots+x\alpha^{k-2}+\alpha^{k-1}).$$
It is the other direction that fails in characteristic 0.