$f'(a)=0$ implies $x=a$ is not a simple zero of $f$ Let $a$ be the root of a polynomial $f(x)$ and let $f'(a)=0$. Then $x=a$ is not a simple  zero of $f(x)$. 
What is the name of this theorem and does someone know a simple (high school level) proof?
 A: I'm not sure the name of it but a simple proof would be:
Since $a$ is a root of $f(x)$ we can write $f(x) = (x-a)g(x)$ for some polynomial $g(x)$. Then taking the derivative we get
$$f'(x) = g(x) + (x-a)g'(x)$$
Now plugging in $a$ we get:
$$0 =f'(a) = g(a) + (a-a)g'(a) = g(a)$$
So $a$ is a root of $g$ and we can write $g(x) = (x-a)h(x)$. Hence $f(x) = (x-a)^2h(x)$ and $a$ is a double root
A: Obviously we are assuming $a$ is a root in the first place.
Suppose $a$ is a single root.
So let $f(x)=(x-a)g(x)$ where $g(a) \ne 0$.
Then $f'(x)=(x-a)g'(x)+g(x)$, so $f'(a)=g(a) \ne 0$, a contradiction.
This proves that $a$ is at least a double root, it could have multiplicity $3$ or higher, though.
A: Consider the product rule of the derivative:
$$\frac {\text d(f(x)g(x))}{\text dx}=f'(x)g(x)+g'(x)f(x)$$
When $f(x)$ has a double root at $a$, which is a double factor, we have $f(x)=(x-a)^2g(x)$ which has derivative $(x-a)^2g'(x)+2(x-a)g(x)$ and thus shows that $f'(a)=0$.
In the other direction, note that $x-a$ must be a factor of $f'(x)$ (by assumption) as $f'(x)=(x-a)h(x)$.  Suppose to the contrary that $f(x)$ does not have a double root at $a$, only a single root so that $f(x)=(x-a)j(x)$ and thus $f'(x)=(x-a)j'(x)+j(x)=(x-a)h(x)$.  But this means that $x-a$ is a factor of $j(x)$, contradicting our supposition that $x-a$ is a single root of $f(x)$ and a single root of $f'(x)$.
