# Factorization of bivariate polynomial

Let $q(y, z) = u_1 + u_2y + u_3 z + u_4y^2 + u_5yz + u_6z^2 + u_7y^3 + u_8y^2z + u_9yz^2 +$ $\hspace{2.55cm}u_{10}y^3z + u_{11}y^2z^2 + u_{12}y^3z^2$

Can $q(y, z)$ be factorized as $$q(y, z) =(v_1+v_2y)(v_3+v_4y+v_5z+v_6yz)(v_7+v_8y+v_9z+v_{10}yz)?$$

Here, {$u$} and {$v$} are complex numbers.

Are there any general principles to factorize a bivariate polynomial?

Thanks.

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Just expand your product and see if it agrees. There are several techniques for factorization, I suggest to look at different CAS. – Per Alexandersson Feb 4 '13 at 9:48
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. – user31145 Feb 4 '13 at 10:16
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! – Bruno Feb 4 '13 at 11:45
you have 12 independent coefficients $u$ and only 10 independent $\nu$'s; so the answer is no. – Carlo Beenakker Feb 4 '13 at 13:24
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. – Dima Pasechnik Feb 4 '13 at 14:20

For a computer algebra implementation, you could try Sage. But you will not find any nontrivial factorization of your polynomial for arbitrary complex coefficients $u_n$.