The correct option is D (x−ω)(x−ω2)
Let f(x)=x6+4x5+3x4+2x3+x+1.
Hence, f(x)=ω6+4ω5+3ω4+2ω3+ω+1
⇒f(x)=1+4ω2+3ω+2+ω+1
⇒f(x)=4(ω2+ω+1)
⇒f(x)=0
Hence, f(x) is divisible by x−ω. Then, f(x) is also divisible by x−ω2 (as complex roots occur in conjugate pairs).
f(−ω)=(−ω)6+(−4ω)5+3(−ω)4+2(−ω)3+(−ω)+1
⇒f(−ω)=ω6−4ω5+3ω4−2ω3−ω+1
⇒f(−ω)=1−4ω2+3ω−2−ω+1
⇒f(−ω)≠0