Partial factorization?

[DrD]

(01-25-2018 01:15 PM)Joe Horn Wrote:  I don't know how Prime can know which "partial factor" you want. Same problem as noting that sqrt(144) = 12, but ifactor(144) factors all the way down. But 12 is in the idivis(144) list. So one approach you can try for your example is to use divis() instead of factor().

I don't know if it is fair play, or not, but one way to know which partial factor is wanted, would be if a command existed that presented a list of all factors.

Since my example was begging for a perfect square to rid itself of the radical, just eyeballing a list of factors would be useful. A command similar to the divis(), but (4*x^4 +1)^2 doesn't directly show up in the resulting vector:

divis(16*x^8+8*x^4+1);

[ 1,
2*x^2-2*x+1,
(2*x^2-2*x+1)^2,
2*x^2+2*x+1,
(2*x^2+2*x+1)*(2*x^2-2*x+1),
(2*x^2+2*x+1)*(2*x^2-2*x+1)^2,
(2*x^2+2*x+1)^2,
(2*x^2+2*x+1)^2*(2*x^2-2*x+1),
(2*x^2+2*x+1)^2*(2*x^2-2*x+1)^2
]

So far anyway, I haven't found a way to do this, directly, on the prime.