Polynomials and Derivatives 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?
 A: If all zeros of $p$ lie in the interval $[a,b]$, then all zeros of $p'$ lie in $[a,b]$ by the Gauss–Lucas theorem.  More generally, each zero of $p'$ lies in the convex hull of the set of zeros of $p$. 
A: 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.
A: Yes. Suppose we have a polynomial $p(x)$ of degree $d$, with all roots real (meaning it has $d$ real roots counting multiplicity). If $p(x)$ has a root of multiplicity $m$ at some point $x_0$, then $p'(x)$ has a root of multiplicity $m-1$ at $x_0$, as can be seen by applying product rule: 
$$\frac{d}{dx}(x-r)^mq(x)=(x-r)^{m-1}(mq(x)+(x-r)q'(x))$$
Also, between any pair of distinct roots of $p(x)$ there must be a root of $p'(x)$ by Rolle's theorem. Thus the number of real roots of $p'(x)$, counting multiplicity, is the number of distinct roots of $p(x)$ minus 1, plus $m-1$ roots for each repeated root of multiplicity $m$, giving a total of $d-1$ (since the number of distinct roots of $p(x)$ plus $m-1$ roots for each repeated root of multiplicity $m$ is the total number of roots of $p(x)$, which is $d$). This is the total number of roots of $p'(x)$, so all its roots must be real.
