Suppose $f$ is an irreducible polynomial and $\xi$ is a root of $f$. If $g$ is any other polynomial such that $g(\xi)=0$, then $f\mid g$. Indeed, if we call $h$ the greatest common divisor of $f$ and $g$, then $h$ is not $1$, since it has $\xi$ a a root. As $h$ is then a non-contant polynomial which divides $f$, it must be a scalar multiple of $f$ itself. Since $h$ divides also $g$, we see that $f$ divides $g$.
Over $K=\Bbb{F}_p$ it does hold that if, for some $f(x)\in K[x]$, we have $f'(x)=0$, then $f$ is reducible. The reason is the following.
If $f'(x)=0$ this means that all the terms in $f(x)$ have degrees that are multiples of $p$. In other words,
$$
f(x)=\sum_{i=0}^na_ix^{pi}
$$
for some natural number $n$ and some coefficients $a_i\in\Bbb{F}_p$.
Two key results then come to the fore:
- In a commutative ring $R$ of characteristic $p$ we have the formula (also known as Freshman's dream) for all $a,b\in R$: $$(a+b)^p=a^p+b^p.$$
- For all $a\in\Bbb{F}_p$ we have $a^p=a$ (Little Fermat).
Put together these imply that the above polynomial $f(x)$
$$
f(x)=\sum_{i=0}^na_ix^{pi}=\sum_{i=0}^n(a_ix^{i})^p=\left(\sum_{i=0}^na_ix^i\right)^p
$$
is actually the $p$th power of a lower degree polynomial, hence reducible.
The same result holds for other fields $K$ of characteristic $p$ as long as all the elements of $K$ are $p$th powers of some element of $K$ (above it would have been enough to have $a_i=b_i^p$ for some $b_i\in K$). Such fields are called perfect, and whenever $K$ is a perfect field we see that irreducible polynomials over $K$ are necessarily separable.
Therefore we can use as the field $K$ any finite field. This is because the Frobenius automorphism $z\mapsto z^p$ is an injective endomorphism of $K$ (trivial kernel). When $K$ is finite "injectivity $\implies$ surjectivity" and we are done.
It doesn't work as nicely for all fields of characteristic $p$. The textbook counterexample is $K=\Bbb{F}_p(t)$. The polynomial $m(T)=T^p-t$ is irreducible (Eisenstein), but it is not separable. It has a single zero $t^{1/p}$ of multiplicity $p$ in an extension field of $K$.
Best Answer
In characteristic $0$, an irreducible polynomial has distinct roots (in a suitable extension field). Indeed, $\gcd(f,f')$ is a divisor of $f$, so it is either $1$ or $f$ (up to multiplication by nonzero constants).
Suppose it is $f$; then $f'=0$, which means $f$ has degree $0$: impossible.
Now, let the characteristic be $p$. If $f$ is inseparable and irreducible, then $f(x)=g(x^p)$, for some polynomial $g$; let $g(x)=a_0+a_1x+\dots+a_nx^n$. Since $F$ is perfect, we have $a_i=b_i^p$, for some $b_i\in F$ $(i=0,1,\dots,n)$. Thus $$ f(x)=\sum_{i=0}^n b_i^px^{ip}=\biggl(\,\sum_{i=0}^n b_ix^i\biggr)^{\!p} $$ contradicting the assumption that $f$ is irreducible.