>
> A 3rd order polynomial can always be factored into a 1st order and a 2nd
> order polynomial. Are you saying that Maxima can't do it?
>
Of course Maxima can do it, but not with the Factor function, which is
factoring over the integers. In general, users don't expect to see factors
in radicals, such as
(x-sqrt(6*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2)-
(3*(sqrt(283)/(6*sqrt(3))+1/2)^(2/3)-4)^(3/2))/
(2*sqrt(3)*(sqrt(283)/(6*sqrt(3))+1/2)^(1/6)*
(3*(sqrt(283)/(6*sqrt(3))+1/2)^(2/3)-4)^(1/4))
+sqrt(3*(sqrt(283)/(6*sqrt(3))+1/2)^(2/3)-4)
/(2*sqrt(3)*(sqrt(283)/(6*sqrt(3))+1/2)^(1/6)))
(one of the real factors of x^4+x-1).
In your case, since you are starting with floating-point coefficients
anyway, an exact factorization in terms of radicals is probably otiose.
Though Maxima specializes in exact symbolic calculations, in this case I'd
think it would be better to carry through the floats.
You can use allroots to calculate the roots numerically, then pair up the
complex roots to give a product of linear and quadratic factors, all with
approximate coefficients, in this case
In the case of non-rational but exact solutions, you can use Solve in a
similar way to (for example) factor x^3-2 into
(x-2^(1/3))*(x^2+2^(1/3)*x+2^(2/3)).