Let $\mathbb{C}[X_1,...,X_d]$ denote the space of polynomials over $\mathbb{C}$ in $d$ variables. In the case $d = 1$, we can always factorize a polynomial using its zeros and their multiplicity, i.e. for any $p \in \mathbb{C}[X]$ we can write it as \begin{align*} p(X) = C (X-a_1) ... (X-a_n) \end{align*} where $C \in \mathbb{C}$, $n \in \mathbb{N}$ is the degree of $p$ and $a_1,...,a_n \in \mathbb{C}$ are the zeros of $p$.
Since $\mathbb{C}[X_1,...,X_d]$ is a unique factorization domain, we can factorize every element $p \in \mathbb{C}[X_1,...,X_d]$ such that \begin{align*} p = C p_1 ... p_n, \end{align*} where $C \in \mathbb{C}$ and $p_i \in \mathbb{C}[X_1,...,X_d]$ are irreducible polynomials. However, since we don't really know any characterization of the irreducible elements of $\mathbb{C}[X_1,...,X_d]$, I don't see how we could make this formula more precise. Is there any additional information we have about the irreducible polynomials $p_i$ based on $p$? For example, do we know anything about the degree of the polynomials?
As an additional question: Suppose we had information about the zeros of $p \in \mathbb{C}[C_1,...,C_d]$, could we then conclude something about the irreducible polynomials $p_i$? For example, suppose we knew that $p(X_1,...,X_d) = 0$ can only hold if there exists an index $i$ such that $X_i = 0$, would we be able to conclude that $p(X_1,...,X_d) = C X_1^{n_1}...X_d^{n_d}$ for $n_i \in \mathbb{N}_0$?