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Question

The polynomial x6+4x5+3x4+2x3+x+1 is divisible by (where ω is one of the imaginary cube roots of unity)


A
x+ω
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B
x+ω2
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C
(x+ω)(x+ω2)
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D
(xω)(xω2)
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Solution

The correct option is D (xω)(xω2)
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 xω. Then, f(x) is also divisible by xω2 (as complex roots occur in conjugate pairs).
f(ω)=(ω)6+(4ω)5+3(ω)4+2(ω)3+(ω)+1
f(ω)=ω64ω5+3ω42ω3ω+1
f(ω)=14ω2+3ω2ω+1
f(ω)0

Mathematics

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