# $f'(a)=0$ then show that $f(a)=0$

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 $$a = 3$$).

• 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. – 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 $$b_1. Then $$\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 $$f(a)=0$$.