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
B
x+ω2
C
(x+ω)(x+ω2)
D
(xω)(xω2)

Solution

## 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)≠0Mathematics

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