Complex numbers and polynomials - University College Cork

Plainly, the pair a = 1, b = 2 works. Hence,X 3 −6X +10 = (X +3)(X +ω+2ω 2 )(X +ω 2 +2ω) = (X +3)(X −1+ω 2 )(X −1+ω).In other words,p(x) = (x + 4)(x + ω 2 )(x + ω)and −4, −ω, −ω 2 are the roots of p.Return to (1). For these relations to hold, a 3 , b 3 must be the roots of the quadraticz 2 − rz − q327 ,and so can be determined by the usual formula,√a 3 , b 3 = r ± r 2 + 4q327.2So, a, b can be determined by extracting cube roots of possibly complex numbers.We could, for instance, choose a to be one of the cube roots of√r + r 2 + 4q327,2something we gloss over.Turning to the determination of the roots of a general cubic p(x) = ax 3 + bx 2 +cx + d, a ≠ 0, since the roots don’t depend on the sign of a, we can suppose, forsimplicity, that a = 1. The first step to perform is to eliminate the term involvingx 2 by shifting x. In fact,x 3 + bx 2 + cx + d = (x + 1 3 b)2 − 1 3 b2 x − b327 + cx + dwhere= (x + 1 3 b)2 + (c − 1 3 b2 )(x + 1 b3b) −3 27 − (c − 1 3 b2 )( 1 3 b) + d= X 3 + (c − 1 3 b2 )X + d − 1 2b3bc +3 27= X 3 + qX + r,X = x + 1 3 b, q = c − 1 3 b2 , r = d − 1 2b3bc +3 27 .Next we select a, b so that−3ab = q, a 3 + b 3 = r;this entails solving the quadratic equation z 2 − rz − q 3 /27 = 0. The final step is touse the factorisation in the above Corollary.17