Complex numbers and polynomials - University College Cork

More generally, the identitiesx 3 + y 3 = (x + y)(x 2 − xy + y 2 ), x 3 − y 3 = (x − y)(x 2 + xy + y 2 ),hold for all real or complex x, y.Recall the definition of ωLemma 2. Suppose x, y, z are real or complex numbers. ThenProof. For,x 2 + y 2 + z 2 − xy − yz − zx = (x + ωy + ω 2 z)(x + ω 2 y + ωz).(x + ωy + ω 2 z)(x + ω 2 y + ωz)= x 2 + xy(ω 2 + ω) + xz(ω + ω 2 ) + ω 3 y 2 + yz(ω 2 + ω 4 ) + ω 3 z 2= x 2 − xy − xz + y 2 + yz(ω 2 + ω) + z 2= x 2 + y 2 + z 2 − xy − yz − zx.Combining these factorisations we deduce that, for all x, y, z,x 3 + y 3 + z 3 − 3xyz = (x + y + z)(x + ωy + ω 2 z)(x + ω 2 y + ωz).□As an immediate consequence, we have the following statement.Corollary 1. The roots of the cubic polynomialp(x) = x 3 − 3abx + a 3 + b 3are given by−(a + b), −(ωa + ω 2 b), −(ω 2 a + ωb).We can exploit this to determine the roots of cubic polynomials of the form x 3 +qx+r,as long as we can find a, b so that−3ab = q, a 3 + b 3 = r. (1)Before discussing the general case, we look at a simple example.Example 5. Determine the roots of p(x) = x 4 + 3x 2 − 3x + 4.Solution. The first step is the eliminate the x 2 term. We can do this by shifting thex-axis. Noting thatp(x) = (x + 1) 3 − 6x + 3 = (x + 1) 3 − 6(x + 1) + 9,its enough to find the roots of X 3 − 6X + 10. So, choose a, b so that2 = ab, a 3 + b 3 = 9.16