Let $f \in \mathbb{R}\left[x\right]$ be a polynomial with real coefficients in one variable $x$ of degree $n > 0$. Assume that $f$ has $n$ real roots (counted with multiplicities).

Let $a \in \mathbb{R}$ be such that the derivative $f'$ of $f$ can be written as $f'\left(x\right) = \left(x-a\right)^2 g\left(x\right)$ for some polynomial $g$. (In other words, $a$ is a double root of $f'$.)

Prove that $f\left(a\right) = 0$ (that is, $a$ is a root of $f$).

I've seen this stated as an exercise with $n = 2016$, but clearly this must be a general fact.

This is not a purely algebraic problem; if we replaced $\mathbb{R}$ by $\mathbb{C}$, then it would become false (for a counterexample, try $n = 3$, $f = x^3 + 2$ and $a = 3$).

  • $\begingroup$ Wait! This is impossible as we know nothing of the zeroth coefficient of $f$. $f(x)$ and $f(x) + c$ will have to same derivative. So if $f(a) = 0$ then $f(a) + c = c$ and we have no way of determining which. $\endgroup$ – fleablood Feb 14 at 2:48

Write $f$ in the form $$f(x)=\prod_{j=1}^r(x-b_j)^{m_j}$$ with exponents $m_j\geq1$ and $b_1<b_2<\ldots< b_r$. Then $$\sum_{j=1}^r m_j =n\ ,$$ and $$f'(x)=\prod_{j=1}^r (x-b_j)^{m_j-1} \>g(x)$$ for some polynomial $g$. According to Rolle's theorem the derivative $f'$ has at least one zero in each open interval $\ ]b_{j-1},b_j[\ $. These $r-1$ zeros have to be zeros of $g$. Since $g$ has degree $${\rm deg}(g)=(n-1)-\sum_{j=1}^r (m_j-1)=r-1$$ these enforced zeros have to be simple. Now we are told that $f'$ has a zero $a$ of order $\geq2$. It follows that this zero has to be one of the $b_j$ given at the outset, hence $f(a)=0$.

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