Factorizing a polynomial with two variables
Hi, I would like to factorize the following polynomial in t, with an integer indeterminate n,
fn(t)=2n((n−1)tn+1)n−1−nn−1t(n−1)(n−2)(tn−1+1)(t+1)n−1
I expect it to have a factor of (t−1)2 and hope that after division the polynomial in t would have positive coefficients.
Is there any way I can verify this by Sage?
I have tried to check that fn(t) has a factor of (t−1)2 by checking that fn(1)=f′n(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×nn−2=nn−1). Did I do the computation in the "wrong way"? If so, what's the proper way to do it?
Thanks!