Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 \begin{equation} q(y, z) =(v_1+v_2y)(v_3+v_4y+v_5z+v_6yz)(v_7+v_8y+v_9z+v_{10}yz)? \end{equation}

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

Are there any general principles to factorize a bivariate polynomial?


share|cite|improve this question
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

1 Answer 1

up vote 1 down vote accepted

I answered your first question in the comments. To answer your second question, yes, these general principles go under the name of "Hensel lifting", see for example these lecture notes.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.