Solving for the roots of a cubic polynomial

[Namir]

(06-30-2024 11:49 PM)Albert Chan Wrote:  Hi, Namir

x^3 + x^2 = 0, but did not get back roots -1, 0, 0

Instead, I get d=0, X1 = X2 = X3 = -1/3

Try its depressed cubic, with x = y - 1/3

y^3 - y/3 + 2/27 = 0

Sure enough, I get d=0, X1 = X2 = X3 = 0

I think the HP author of the program assumed that parameter c should not be 0, because in this case the cubic equation s easily reduced to a quadratic or linear equation (as is your example).