If x=1+ix=1+i is a factor of ax^2+bx+cax2+bx+c then what is another factor?

1 Answer
May 13, 2016

c) 1-i1i

Explanation:

I will assume that by 'factor' you intended 'zero' or 'root'.

A 'factor' would be something like (x - 1 - i)(x1i), which corresponds to a 'zero' x=1+ix=1+i of ax^2+bx+cax2+bx+c, also called a 'root' of ax^2+bx+c = 0ax2+bx+c=0.

If x = 1+ix=1+i is a zero of ax^2+bx+cax2+bx+c and aa, bb and cc are Real numbers then x = 1-ix=1i is the other zero.

If aa, bb and cc can be Complex numbers, then the other zero can be anything.

In general, given any polynomial with Real coefficients, any zeros are Real or occur in Complex conjugate pairs.

From the perspective of the Real numbers, there is no way to tell the difference between ii and -ii. The only thing the Real numbers 'know' about +-i±i is that they are square roots of -11.