Question

If ω is the complex cube root of unity, then which among the following is a factor of the polynomial x^6+ 4x^5+3x^4+2x^3+x+1?

(a) x+ω

(b) x+ω^2

(c) (x+ω)(x+ω^2)

(d) (x–ω)(x–ω^2)

I got this question by my college director while I was bunking the class.

Enquiry is from Roots of Complex Numbers in division Complex Numbers of Engineering Mathematics

Answer 1

Right choice is (d) (x–ω)(x–ω^2)

Explanation: Let f(x)=x^6+ 4x^5+3x^4+2x^3+x+1. Therefore, f(ω)= ω^6+4ω^5+3ω^4+2ω^3+ω+1=0

Since, ω^2 is the complex conjugate of ω, therefore ω^2 is also a root (roots occur in conjugate pairs). Therefore (x–ω)(x–ω^2) is a root. Also, f(-ω)=(-ω)^6+4(-ω)^5+3(-ω)^4+2(-ω)^3+(-ω)+1=1-4ω^2+3ω-2-ω+1≠0⇒-ω and hence -ω^2 are not the roots.