I know that $x^4-x \in \mathbb{C}[x]$ splits completely and is separable, but I'm not sure what the answer is for $(\mathbb{Z}/2\mathbb{Z})[x]$. It seems that it should obviously be that the polynomial doesn't split and isn't separable. But since only $0,1 \in \mathbb{Z}/2\mathbb{Z}$, and these are roots of the polynomial, I'm not sure what the answer is.
Also, in the definition of splitting (for a polynomial), is the only difference for the definition is separability that we allow the constants to be nondistinct?
This factors as $$x(x^3-1)=x(x-1)(x^2+x+1)$$ But the latter factor is irreducible. The polynomial is separable though. $x^2+x+1$ doesn't have a repeated root because it can't be written as $(x-a)^2$, and neither root is equal to $0$ or $1$.