Repeated roots of a function I have come across questions which states that the functions has repeated roots. There $\alpha$ is said to be  repeated root of $f(x)$ if $f(\alpha)=f'(\alpha)=0$.
If $f(x)$ is a polynomial I can understand why they call it as a repeated root because $(x-\alpha
)^2$ is a factor of $f(x)$. But if $f(x)$ is not a polynomial then you cannot factorise $f(x)$, so can someone tell me the reason why $\alpha$ called a repeated root in such cases. Thanks in advance.
 A: If you can write the function as a Taylor series around $a,$ you can the see the first non-zero term is $a_n(x-a)^n,$ for some $n\geq 2,$ and you can factor out $(x-a)^2.$ (Or possibly, all terms zero, if the function is the zero function.)
More generally, if $f’’(x)$ exists, then it means that you can write $f(x)=(x-a)^2 g(x)$ for some $g(x)$ continuous and defined at $a.$ So again, you can factor out $(x-a)^2.$
You you can get the even stronger result that if $f’’$ exists, then $g$ exists if and only if $f(a)=f’(a)=0.$
This follow by induction from the result:

If $h'$ exists, then $g_1(x)=\frac{h(x)-h(a)}{x-a}$ can be made continuous only by defining $g_1(a)=h'(a).$

That is practically the definition of $h'(a),$ which is $$h'(a)=\lim_{x\to a} g_1(x).$$
Then if $f(a)=f'(a)=0,$ first apply this to $h=f',$ then to $h=f.$
A: I think that the terminology "repeated root" is properly referred to the Taylor polynomial of $f$ centered at $\alpha$ of degree $n$, i.e. $$T_{n,\alpha}(x)=\sum_{k=0}^{n} \frac{f^{(k)}(\alpha)}{k!}(x-\alpha)^k.$$
Indeed, if $f(\alpha)=f'(\alpha)=0$ then the polynomial $T_{n,\alpha}$ is divisible by $(x-\alpha)^2$.
