Maple

Hi!

I a have a question about factorizing real polynomials.

Suppose I have a real polynomial p(x) with integer coefficients. If the degree of p(x) is less than or equal to 4, we can factorize it into linear radical factors. On the other hand, if we require the factorization to be real, theoretically we can factorize it into linear and irreducible quadratic factors.

My question is, if the input p(x) is real polynomial with integer coefficients, is there any Maple function that can give me factorization output with real linear and irreducible quadratic factors, with radical coeffs?

For example, I tried q := 20*x^3+10*x^2+4*x+1, it has one real root and 2 complex roots. I want a factorization like q(x) = 20*(x-r1)*(a*x^2 + b*x + c), with r1, a, b, c, all real radicals.

I compared functions: factors(), solve(), sqrfree(), Splits(), and none of them give what I want.

factors(q) gives:
[20, [[x^3+(1/2)*x^2+(1/5)*x+1/20, 1]]]

factors(q, real)  gives:
[20., [[x+.3423840948583691316993036540027816871936619136844427977504078911, 1], [x^2+.1576159051416308683006963459972183128063380863155572022495921089*x+.1460348209828001458360112632660894203743660942160039146818509889, 1]]]

solve(q)   gives:
-(1/30)*(350+105*sqrt(15))^(1/3)+7/(6*(350+105*sqrt(15))^(1/3))-1/6, (1/60)*(350+105*sqrt(15))^(1/3)-7/(12*(350+105*sqrt(15))^(1/3))-1/6+(1/2*I)*sqrt(3)*(-(1/30)*(350+105*sqrt(15))^(1/3)-7/(6*(350+105*sqrt(15))^(1/3))), (1/60)*(350+105*sqrt(15))^(1/3)-7/(12*(350+105*sqrt(15))^(1/3))-1/6-(1/2*I)*sqrt(3)*(-(1/30)*(350+105*sqrt(15))^(1/3)-7/(6*(350+105*sqrt(15))^(1/3)))