A-Level Further MathsFactorising polynomials with complex rootsOn this pageFactorising polynomials with complex roots Factor theorem Using the factor theorem, we know that if aaa is a factor of f(x)f(x)f(x), one of the factors of f(x)f(x)f(x) is x−ax-ax−a. This also works for polynomials with complex roots. If a+bia+bia+bi is a root, that means that (x−[a+bi])(x-[a+bi])(x−[a+bi]), otherwise written as (x−a−bi)(x-a-bi)(x−a−bi), is a factor. Factorise x2−4x+40x^2-4x+40x2−4x+40 by first solving Let x2−4x+40=0x^2-4x+40=0x2−4x+40=0 (x−2)2−4+40=0(x-2)^2-4+40=0(x−2)2−4+40=0 (x−2)2=−36(x-2)^2=-36(x−2)2=−36 x−2=−36x-2=\sqrt{-36}x−2=−36 x=2±6ix=2\pm6ix=2±6i Roots: x=2+6ix=2+6ix=2+6i or x=2−6ix=2-6ix=2−6i Factors (by the factor theorem): [x−(2+6i)][x-(2+6i)][x−(2+6i)] and [x−(2−6i)][x-(2-6i)][x−(2−6i)]