MaplePrimes - answers and comments on Question, Does maple ALWAYS find the factorization of a polynomial?

factorization ~ zeroes

Axel Vogt — Sun, 04 Dec 2011 15:58:29 Z

Usually factorization into linear or quadratic facors is meant.

But that is equivalent to know the real or complex-paired zeros.

If you additionally allow for parameters a[j], then it means that
you take an arbitrary polynomial and ask for its zeroes.

But you can not expect that Maple will do it. Or do you?

Factoring over a field

Robert Israel — Mon, 05 Dec 2011 05:01:20 Z

@edgar : By default, Maple looks for a factorization over the field generated by the coefficients. If you want a factorization over some algebraic number field, you can provide that as a second argument to factor, e.g.

> factor(x^2 - 2, sqrt(2));

For a univariate polynomial, you can also ask for a complete factorization over the (floating-point) real or complex field, e.g.

> factor(x^5 - 3*x + 2, real);

> factor(x^5 - 3*x + 2, complex);

thanks for the reply. for example, f:=x^2+(a+b

LijiH — Sun, 04 Dec 2011 23:39:40 Z

thanks for the reply.

for example, f:=x^2+(a+b)*x+(a*b)

I would like Maple to factor this into (x+a)*(x+b). In this case, Maple indeed manages to factor it without a problem, but I was wondering what if my polynomial is quite long and has lots of parameters, is Maple always "reliable"?

parameters

Robert Israel — Mon, 05 Dec 2011 05:08:46 Z

@LijiH : the "parameters" are considered as variables, on the same footing as x. Maple is indeed quite reliable at finding whatever factorization is possible (into irreducible multivariate polynomials over the field generated by the coefficients). For example:

> factor(x^4 - 2*a*x^2 + a^2);

If it is really long and has lots of variables, that may slow Maple down somewhat (although the algorithms are pretty efficient), but shouldn't affect the reliability.