# Finding roots and factors of multivariate polynomials

I know that in order to factor a one dimensional polynomial one can find the roots with some method, for instance a numerical newton method. Then one can systematically divide with $(variable-root)$ for each root found and then be done. Is there any analogous way to do this for multivariate polynomials? Does there exist any "unique" or "natural" factorization for those? It is obvious we can do this in the case our polynomial is separable i.e. $$P(x_1,x_2,\cdots,x_n) = P_1(x_1)P_2(x_2) \cdots P_n(x_n)$$ because then we could just factor each $P_k(x_k)$ separately.

But what about the general case?

• You can apply Newton-Raphson for multivariate roots. – MrYouMath Sep 16 '15 at 9:56
• So will I be able to factor them with using just about any root which I find with Newton-Raphson? – mathreadler Sep 16 '15 at 9:57
• Lets say you focus on $x_1$ first. Then think of all the other variables as "constants". Then you find one root $x_11$. Now devide by $(x_1-x_11)$. I don't see any problem here. But what you might get is that the coefficients of your remaining polynomial become very complicated. You should just try. – MrYouMath Sep 16 '15 at 10:02
• No, it is in general not possible to factor multivariate polynomials over the real or complex numbers. In general, the generic multivariate polynomial is irreducible. -- @MrYouMath: the root $x_{1,1}$ is an algebraic function in the other variables, in general it will not be a constant. – LutzL Sep 16 '15 at 22:15
• Besides the general case, there is a class of symmetric polynomials that describe toric varieties. They have amazing multivariate factoring properties. Extremely natural, you could say. – Philip Pugeau Nov 22 '15 at 4:39