= Solution
<Rolle theorem> states that if $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is $c\in(a,b)$ with $f'(c)=0$.
If $f$ is constant, any interior point works. Otherwise the <extreme value theorem> gives an attained maximum and minimum on the interval, and at least one of them differs from the common endpoint value. That extremum must occur at an interior point $c$. At an interior maximum, for small positive $h$ the difference quotient $[f(c+h)-f(c)]/h$ is nonpositive, while for negative $h$ it is nonnegative. Existence of the common derivative forces it to be both nonpositive and nonnegative, hence zero. The same argument with inequalities reversed applies to an interior minimum. This proves <Rolle theorem>, including the derivative-at-an-extremum step rather than assuming it.
Now let the nonconstant <polynomial> $p$ have distinct real roots $r_1<\cdots<r_k$, with <multiplicities> $m_1,\ldots,m_k$ adding to its degree $n$. If $p(x)=(x-r_j)^{m_j}q_j(x)$ and $q_j(r_j)\ne0$, differentiation gives
$$
p'(x)=(x-r_j)^{m_j-1}\bigl[m_jq_j(x)+(x-r_j)q_j'(x)\bigr].
$$
The bracket is nonzero at $r_j$, so its <multiplicity> as a root of $p'$ is exactly $m_j-1$. These roots already contribute $\sum_j(m_j-1)=n-k$ real roots counted with multiplicity. <Rolle theorem> supplies an additional root of $p'$ in each of the $k-1$ disjoint intervals $(r_j,r_{j+1})$. These additional roots are distinct from each other and from all the $r_j$.
We have therefore exhibited at least $(n-k)+(k-1)=n-1$ real roots counted with multiplicity. The derivative has degree $n-1$, so there is no room for any additional nonreal roots. \b[Every root of $p'$ is real.] This is the <Rolle root count with multiplicities>; for $n=1$ the nonzero constant derivative has no roots, as expected.
For failure of the converse, take
$$
\boxed{p(x)=x^3+1,\qquad p'(x)=3x^2.}
$$
The derivative has only the real root zero, of multiplicity two. But $p(x)=(x+1)(x^2-x+1)$ and the quadratic factor has roots $(1\pm i\sqrt3)/2$, which are nonreal. Thus real-rootedness of the derivative does not imply real-rootedness of the original <polynomial>.
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