You use Rolle's theorem to prove n-polynomial has n distinct roots, but how would you do this question? Show that if a polynomial of degree n has n real roots (which don't have to be distinct) then its derivative has n-1 real roots (which also don't have to be distinct).
Let us consider the distinct roots $s_1,s_2,\ldots s_{\ell}$ with respective multiplicities $b_1,b_2,\ldots b_{\ell}$. Now see that $b_1+b_2+\ldots+b_{\ell}=n$ now the $\ell$ distinct roots give $\ell-1$ roots of the derivative and the multiplicities contribute $b_i-1$ each. Adding them up we get $\left(\displaystyle \sum_{i=1}^{\ell}b_i\right)+\ell-1=n-1$ roots.
$\blacksquare$
Suppose that $p(x)$ is a polynomial with $n$ ordered real roots, $a_1 \leq ...,\leq a_n$.  Then since $p$ is continuous on $[a_1, a_2]$, differentiable on $(a_1, a_2)$, and $p(a_1) = p(a_2)$ we can apply Rolle's theorem to conclude that there exists a real number, $c_1 \in (a_1, a_2)$ such that $f^\prime(c_1) = 0$.  Repeat this argument for the $n-2$ remaining intervals $(a_j, a_{j+1})$.