TITLE: Polynomials and Derivatives
QUESTION [10 upvotes]: I am doing an exercise and came to a point where I'd need to know this:
As a consequence of Rolle's theorem, the derivative of a function has a zero whenever our function has more than one zero. But can we say that if all the roots of a polynomial, $p(n)$, are real, all the roots of $p'(n)$ are real?

REPLY [2 votes]: HINT $\rm\ \qquad \ f\ =\ (x-r_1)^{n_1}\:\cdots\:(x-r_k)^{n_k},\qquad\qquad \rm r_1 < r_2 <\cdots < r_k \in \mathbb R$ 
$\rm\qquad\quad \Rightarrow\: \qquad f\:\:' =\: (x-r_1)^{n_1-1}\:\cdots\ (x-r_k)^{n_k-1}\: g\quad for\ some\ g\in \mathbb R[x] $ 
$\rm\qquad\quad \Rightarrow\ \ \ \#roots\ f\:\:' \ge\: (n_1\!\!-1)+\cdots+(n_k\!\!-1)\ +\ k\!-\!1\ \ \ by\ f\:\:'\ has\ root\ in\ (r_i,r_{i+1})\ by\ Rolle$
$\rm\phantom{\rm\ \qquad\quad \Rightarrow\ \ \#roots\ f'} \ge\:  deg\ f - 1$
$\rm But\ also\:\:\ \ \#roots\ f\:\:' \le deg\ f - 1\ =\ deg\ f\:\:'\ $ since $\rm\:\mathbb R\:$  is a domain.