This question might not be interesting but I still want to know the answer either way.
The Gauss Lucas theorem on complex number $C$ says the roots of $\frac{dp(x)}{dx}$ for $p\in C[x]$ lies in the convex hull formed by roots of $p(x)$.
It seems this holds over $Q$ under the assumption that $p'(x)$ has at least one solution over $Q$.
Apparently, I cannot deal with derivative of a polynomial in general over characteristic $p$ field. Consider field $F$ with $char(F)=p$ and its associated algebraic closure $\bar{F}$. Suppose $p\not\in k[x^p]$. I think above thm with $C$ replaced by $\bar{F}$ will hold for $p\not\in k[x^p]$'s case either way. Similarly suppose $p'(x)$ has a solution over $F$, then the thm holds for $F$ replacing $C$ as well. Is this correct? Do I need some extra structure?
No, this doesn't even make sense. "Convex hull" is meaningless in a more general context: to make sense of it, you need to looking at something like a subset of a vector space over an ordered field, so you can talk about linear combinations with coefficients in $[0,1]$.
(For that matter, no such result is true over $\mathbb{Q}$ either without stronger hypotheses--maybe you want to assume $p$ splits over $\mathbb{Q}$? For instance, taking $p(x)=x^2+1$, then $p$ has no roots over $\mathbb{Q}$ but $p'$ does.)