Question

The polynomial $x^6+4x^5+3x^4+2x^3+x+1$ is divisible by- (where $\omega$ is complex cube root 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 $(x-\omega )(x-{\omega }^2)$

$Thepolynomial:x^6+{4x}^5+3x^4+2x^3+x+1letthecuberootofunitybe{x}^2+x+1\therefore Factorsare:(x^4+3x^3-x^2+1)(x^2+x+1)x^2+x+1hasrootsw\&w^2\therefore x^2+x+1=(x-w)(x-w^2)\therefore polynomialhasfactors(x^4+3x^3-x^2+1)(x-w)(x-w^2)$