The polynomial `x^6+4x^5+3x 64+2x^3+x+1` is divisible by_______ where `w` is the cube root of units `x+omega` b. `x+omega^2` c. `(x+omega)(x+omega^2)` d. `(x-omega)(x-omega^2)` where `omega` is one of the imaginary cube roots of unity.

Options

A. `x + omega `
B. ` x + omega^(2)`
C. `( x+ omega) (x + omega^(2))`
D. `(x + omega)(x - omega^(2))`

Answer

D

Explanation

Let `f(x) = x^(6) + 4x^(5) + 3x^(4) + 2x^(3) + x+ 1`. Hence,
`f(omega) = omega^(6) + 4omega^(5) + 3omega^(4) + 2omega^(3) + omega + 1`
`= 1+ 4 omega ^(2)+ 3omega^(4) + 3omega + 2 + omega +1`
`= 4(omega^(2) + omega + 1)`
`= 0`
Hence, `f(x)` is divisible by `x-omega` Then `f(x)` is also divisible by `x - omega^(2)` (as complex roots occur in conjugate pairs).
`f(-omega) = (-omega)^(6) + 4(-omeaga)^(5) + 3(-omega)^(4) + 2(-omega)^(3) + (-omega) +1`
`= omega^(6) - 4omega^(5) + 3omega^(4) - 2omega^(3) - omega + 1`
` = 1 - 4omega^(2) + 3omega- 2 - omega + 1`
` ne 0`.