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Factorizing a polynomial with two variables

asked 12 years ago

Sanchez gravatar image

updated 12 years ago

Hi, I would like to factorize the following polynomial in t, with an integer indeterminate n,

fn(t)=2n((n1)tn+1)n1nn1t(n1)(n2)(tn1+1)(t+1)n1

I expect it to have a factor of (t1)2 and hope that after division the polynomial in t would have positive coefficients.

  1. Is there any way I can verify this by Sage?

  2. I have tried to check that fn(t) has a factor of (t1)2 by checking that fn(1)=fn(1)=0. I named my polynomial f, and let h be f.derivative(t), and try to find h(t=1). Here's what I got,

(n - 1)^22^nnn^(n - 2) - (n - 1)2^(n - 1)n^(n - 1) - 2(n - 1)2^(n - 2)n^(n - 1) - 2(n^2 - 3n + 2)2^(n - 1)n^(n - 1)

Which turns out to be 0 when I verify by hand. However it seems like Sage is unable to detect the redundance in the expression (e.g. n×nn2=nn1). Did I do the computation in the "wrong way"? If so, what's the proper way to do it?

Thanks!

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answered 12 years ago

ppurka gravatar image

Try to use full_simplify on the expression

sage: var('n')
n
sage: implicit_multiplication(True)
sage: F = (n - 1)^2 2^n n n^(n - 2) - (n - 1)2^(n - 1)n^(n - 1) - 2(n - 1)2^(n - 2)n^(n - 1) - 2(n^2 - 3n + 2)2^(n - 1)n^(n - 1)
sage: F.full_simplify()
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Thanks! That settles my second question.

Sanchez gravatar imageSanchez ( 12 years ago )

For the first question, I think your method is correct. There can be some other methods, like looking at the factors of f(t) using `f.factor_list()`, or by looking manually at the output of `f(t)/(t-1)^2`, etc. But these are going to be slow and not always give results which are in their simplest form.

ppurka gravatar imageppurka ( 12 years ago )

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Asked: 12 years ago

Seen: 449 times

Last updated: Jan 21 '13