Starting by example: If for an analytic function $f$ one has:
$$ f(x_1) = 0, \quad f'(x_1) = 0, \quad f(x_2) = 0 $$
can $f$, in its most general form, be always written as
$$ f(x) = (x-x_1)^2(x-2)g(x) $$
where $g$ is (some) member of the set of all analytic functions?
Generally speaking, if for analytic $f$ one has
$$ f^{(k)}(x_n) = 0 \text{ for all } k<k_n $$
(for somehow chosen $k_n$) can then $f$ be always written as
$$ f(x) = g(x)\times \Pi_{i=1}^{n_{max}} (x-x_i)^{k_i} $$
where $g$ is a general analytic function. ?
(I suppose you are talking about analytic functions on the entire complex plane).
This is true and it follows from the fact that if $f$ has a zero of order $n$ at $z_0$ then $f(z)=(z-z_0)^{n}g(z)$ for some analytic function $g$.
Proof: Let $f(z)=\sum\limits_{k=0}^{\infty} a_k (z-z_0)^{k}$ be the power series expansion of $f$. If $f(z_0)=0$ and $f$ is not identically $0$ then there is a smallest integer $n$ such that $a_0=a_1=...=a_{n-1}=0$ and $a_n \neq 0$. This $n$ is called the order of zero of $f$ at $z_0$. Since the original power series converges so does $\sum\limits_{k=n}^{\infty} a_k(z-z_0)^{k-n}$. [The two series have the same radius of convergence. Let $g(z)$ be the sum of this series. Then $f(z)=(z-z_0)^{n} g(z)$ and $g$ is analytic.