Necessary condition for a polynomial of several complex variables to vanish at a point

Question

For a $1$-variable polynomial $f(x)\in\Bbb C[x]$ it is well-known that $$f(a)=0\iff f(x)=(x-a)g(x)$$ for some polynomial $g(x)\in\Bbb C[x]$.

Question: Does that principle carries over to multivariable polynomials?

That is if $f(x_1,\ldots,x_n)\in\Bbb C[x_1,\ldots,x_n]$ and $f(a_1,\ldots,a_n)=0$ for some $(a_1,\ldots,a_n)\in\Bbb C$, does it follow that $$f(x_1,\ldots,x_n)=(x_1-a_1)\cdots(x_n-a_n)g(x_1,\ldots,x_n),$$ for some $g\in\Bbb C[x_1,\ldots,x_n]$?

Attempt: We can fix $x_2,\ldots,x_n$ and consider $f(x_1,\ldots,x_n)$ as a $1$-variable polynomial in $x_1$ that vanishes at $a_1$ and hence $f(x_1,\ldots,x_n)=(x_1-a_1)g(x_1)$, but I am not sure why $g(x_1)$ would be a polynomial in $x_2,\ldots,x_n$.

Answer 1

Consider the polynomial $f(x_1, x_2) = x_1 + x_2$. Although $f(0, 0) = 0$, $f(x_1, x_2) \neq x_1x_2g(x_1, x_2)$ for any polynomial $g$.

Answer 2

Consider $$ f(x_1,x_2) = 1-x_1^2 + x_2^2 - 2x_2 $$ $f(\frac12,\frac12) = 0$ yet $f(x_1,x_2)$ neither has a factor of $(x_1-\frac12)$ nor $(x_2-\frac12)$

Answer 3

As shown by Micheal Albanese and Mark Fischler, there are counterexamples to your tentative generalization. However, let me point out that there is a natural generalization of the one variable case:

If $f(x_1,\ldots,x_n)\in\Bbb C[x_1,\ldots,x_n]$ and $f(a_1,\ldots,a_n)=0$ for some $(a_1,\ldots,a_n)\in\Bbb C^n$, then $$f(x_1,\ldots,x_n)=(x_1-a_1)g_1(x_1,\ldots,x_n)+\cdots+(x_n-a_n)g_n(x_1,\ldots,x_n),$$ for some polynomials $g_i\in\Bbb C[x_1,\ldots,x_n]$.

The proof goes by applying the division algorithm repeatedly for each variables and then evaluating at $(a_1,\ldots,a_n)$ to fix the constant.