Question

The polynomial $x^6+4x^5+3x^4+2x^3+x+1$ is divisible by (where $\omega$ is one of the imaginary cube roots of unity)

• A. $x+\omega$
• B. $x+{\omega }^2$
• C. $(x+\omega )(x+{\omega }^2)$
• D. $(x-\omega )(x-{\omega }^2)$

Answer 1

The correct option is D $(x-\omega )(x-{\omega }^2)$

Let $f\left(x\right)=x^6+4x^5+3x^4+2x^3+x+1$.
Hence, $f\left(x\right)={\omega }^6+4{\omega }^5+3{\omega }^4+2{\omega }^3+\omega +1$
$\Rightarrow f\left(x\right)=1+4{\omega }^2+3\omega +2+\omega +1$
$\Rightarrow f\left(x\right)=4({\omega }^2+\omega +1)$
$\Rightarrow f\left(x\right)=0$
Hence, $f\left(x\right)$ is divisible by $x-\omega$. Then, $f\left(x\right)$ is also divisible by $x-{\omega }^2$ (as complex roots occur in conjugate pairs).
$f(-\omega )=(-\omega {)}^6+(-4\omega {)}^5+3(-\omega {)}^4+2(-\omega {)}^3+(-\omega )+1$
$\Rightarrow f(-\omega )={\omega }^6-4{\omega }^5+3{\omega }^4-2{\omega }^3-\omega +1$
$\Rightarrow f(-\omega )=1-4{\omega }^2+3\omega -2-\omega +1$
$\Rightarrow f(-\omega )\ne 0$