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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?

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    $\begingroup$ You can apply Newton-Raphson for multivariate roots. $\endgroup$
    – MrYouMath
    Sep 16, 2015 at 9:56
  • $\begingroup$ So will I be able to factor them with using just about any root which I find with Newton-Raphson? $\endgroup$ Sep 16, 2015 at 9:57
  • $\begingroup$ 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. $\endgroup$
    – MrYouMath
    Sep 16, 2015 at 10:02
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    $\begingroup$ 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. $\endgroup$ Sep 16, 2015 at 22:15
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    $\begingroup$ 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. $\endgroup$ Nov 22, 2015 at 4:39

1 Answer 1

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The factorization of multivariate polynomials is quite a difficult topic. In brief, the polynomial is converted into a univariate one (one in one variable), by selecting suitable values for the other variables.

The resulting univariate polynomial is then factorized - there are several methods to do this in the general case. The factored polynomial is then 'built up' by using something called Hensel lifting. This is quite an advanced subject area and would typically be taught sort of on 3rd year abstract algebra university courses.

The multivariate factorization problems you get in school or whatever tend to be fairly simple 'toy problems' that have nice neat solutions, by usually employing something like difference of two cubes, or by exploiting some sort of substitution.

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  • $\begingroup$ Did I accept this answer..? What was I thinking? $\endgroup$ May 23, 2020 at 12:24
  • $\begingroup$ Probably because the other answers were wrong? You don't use Newton Raphson to find factors, its used to find roots. $\endgroup$ Nov 6, 2020 at 20:02

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