Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know how to factorize the following expression:

$$4x^4+12x^{10/3} y^{2/3}+33x^{8/3} y^{4/3}+46x^2 y^2+33x^{4/3} y^{8/3}+12x^{2/3} y^{10/3}+4 y^4$$ ?

share|improve this question
    
I removed some tags, changed one, added one, changed title and formatting of question. –  Aryabhata Dec 29 '10 at 6:11
2  
@Mo: You know how it is, some expressions are just too damn big to be placed in question titles... :D –  J. M. Dec 29 '10 at 6:18
    
@J.M: Yup ... :-) –  Aryabhata Dec 29 '10 at 6:19

1 Answer 1

up vote 8 down vote accepted

It is possible, but likely messy.

First set $\displaystyle z = \left(\frac{x}{y}\right)^{2/3}$ and divide your expression by $\displaystyle y^4$.

We get

$$4z^6 + 12 z^5 + 33 z^4 + 46 z^3 + 33 z^2 + 12 z + 4$$

Now divide this by $\displaystyle z^3$ and set $\displaystyle t = z + 1/z$.

We get

$$4(z^3 + 1/z^3) + 12(z^2 + 1/z^2) + 33(z+1/z) + 46$$

$$ = 4(t^3 - 3t) + 12(t^2 - 2) + 33t + 46$$

$$ = 4t^3 + 12t^2 + 21t + 22$$

Which is a messy cubic, according to Wolfram Alpha.

(Note it is always possible to factorize a cubic in "closed form", because of Cardano's method of finding the roots).

share|improve this answer

Your Answer

 
discard

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.