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I am trying to prove that if $f$ is a polynomial where all of its roots are real, then all of the roots of $f'$ are also real. I know that the result holds immediately from Rolle's Theorem, however I cannot use it because it is a result of Calculus. This exercise was spent in an Algebra course. I tried to prove by induction in the degree of $f$, but I was unable to complete the induction step. Can anyone give me a tip?

Rodolfo Ferreira de Oliveira
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    Doesn't the answer to this one give you enough? https://math.stackexchange.com/questions/1901992/is-there-an-algebraic-way-to-prove-this-relationship-between-the-roots-of-a-real?rq=1 – ancient mathematician Apr 04 '21 at 10:50

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Suppose for contradiction that all the roots of $f$ are real but there is a root $\alpha$ of $f'$ which is not real. Let $x_1,\dots,x_n$ be the roots of $f$. We have

$$ \begin{align*} \frac{f'(x)}{f(x)} &= \sum_{i=1}^{n} \frac{1}{x-x_i} \\ 0 = \frac{f'(\alpha)}{f(\alpha)} &= \sum_{i=1}^{n} \frac{1}{\alpha-x_i} \\ &= \sum_{i=1}^{n} \frac{\overline{\alpha}-x_i}{|\alpha-x_i|^2} \\ &=\sum_{i=1}^n \frac{\overline{\alpha}}{|\alpha-x_i|^2} - \sum_{i=1}^n \frac{x_i}{|\alpha-x_i|^2} \end{align*} $$ Solving for $\overline{\alpha}$, we have

$$ \overline{\alpha} = \frac{\sum_{i=1}^n \frac{x_i}{|\alpha-x_i|^2}}{\sum_{i=1}^n \frac{1}{|\alpha-x_i|^2}}, $$ which is real. Contradiction

fwd
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    You need to remove any multiple roots from $f$ before applying this technique, so that you aren't dividing by zero. – C Monsour Apr 04 '21 at 11:40
  • @CMonsour $\alpha$ is not a multiple root of $f$ because it is assumed to be complex. – fwd Apr 04 '21 at 11:46
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The roots of f'(x) lies in the convex hull of the roots of f(x). Since the roots of f(x) are real, the convex hull turns out to be a subset of the real line and hence the roots of f'(x) are real.

Lawrence Mano
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