# Relationship between f and f' in terms of number of real/non-real roots

$f$ is a polynomial of degree $n\ge1$ and $\forall x,x\in \Bbb R \rightarrow f(x)\in\Bbb R$.

Prove that: (a)$f$ has at most one more real root than $f'$ (b)$f'$ has no more non-real roots than $f$ (c)If all roots of $f$ are real, then all roots of $f'$ are real.

I notice that (a) and (b) are equivalent, so if I can prove (a), (b) is automatically proven (because $f'$ has exactly n-1 roots). But I am just totally unaware of the way to start.

Based on the homework context, I think it has something to do with the following: Theorem: Let $z_0$ be a root of multiplicity m of polynomial f: (1)If m=1, then f'($z_0)\neq0$. (2)If m>1, then $z_0$ is a root of f', of multiplicity m-1.

• Please clarify that coefficients are "Real" or "Complex"? – Hoseyn Heydari Feb 7 '14 at 18:29

Between any two consecutive distinct roots of $f$ there is at least one root of $f'$ (Rolle theorem). This shows the claim if we do not count multiplicities. And what happens at a (multiple) root of $f$ is the subject of the theorem yuo quote. Both statements can be combined as follows if you like: If $a_1\le a_2\le \ldots \le a_k$ are the real roots of $f$ (with repetitions for multiple roots) then there is a sequence $b_1\le b_2\le \ldots \le b_{k-1}$ of roots of $f'$ (with repititions possible for multiple roots; note that in the context of complex roots for $f$ there may be additional real roots of $f'$) such that $a_1\le b_1\le a_2\le b_2\le a_3\le \ldots\le a_{k-1}\le b_{k-1}\le a_k$; in the end we also obtain the claim with counting multiplicities.