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Feb 8, 2013 at 4:26 vote accept user31145
Feb 4, 2013 at 20:35 answer added Carlo Beenakker timeline score: 1
Feb 4, 2013 at 17:14 comment added user31145 Can you suggest any other factorization which has more than 12 coefficients?
Feb 4, 2013 at 17:13 history edited user31145 CC BY-SA 3.0
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Feb 4, 2013 at 14:28 comment added Pietro Majer Of course, in case you can take each $u_k$ as a function of the $v_j$, just expand the LHS, equate the coefficients, and get $u_1:=v_1v_3v_7,\dots,u_{12}:=v_2v_6v_{10}$.
Feb 4, 2013 at 14:20 comment added Dima Pasechnik well, there are branches of mathematics, commutative algebra, and algebraic geometry, which, among other, deal with questions like this. In general, bivariate polynomials (with coefficients in an infinite field) almost never factor.
Feb 4, 2013 at 13:24 comment added Carlo Beenakker you have 12 independent coefficients $u$ and only 10 independent $\nu$'s; so the answer is no.
Feb 4, 2013 at 11:45 comment added Bruno In which field/ring do the coefficients of your polynomial lie? Are you looking for a factorization in this field or in an algebraic closure (a.k.a. absolute factorization)? Your question is not really well-defined!
Feb 4, 2013 at 10:28 history edited user31145 CC BY-SA 3.0
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Feb 4, 2013 at 10:16 comment added user31145 I don't have the numerical values of {u} and {v}. I am interested in the terms in each factor. For example, should I add more higher-power terms in certain factors, can I reduce lower-power in some factors, is the factorized form the same as the expanded form for a given factorization? Thanks.
Feb 4, 2013 at 9:48 comment added Per Alexandersson Just expand your product and see if it agrees. There are several techniques for factorization, I suggest to look at different CAS.
Feb 4, 2013 at 9:37 history asked user31145 CC BY-SA 3.0