Question

The polynomial $x^6+4x^5+3x^4+2x^3+x+1$ is divisible by _____ where $w$ is the cube root of units

• A. $x+w$
• B. $x+w^2$
• C. $(x+w)(x+w^2)$
• D. $(x-w)(x-w^2)$

Answer 1

The correct option is D $(x-w)(x-w^2)$

$f\left(x\right)=x^6+4x^5+3x^4+2x^3+x+1$
Since $w$ is the cube root of unity.
$\therefore w^3=1\&1+w+w^2=0$
$f\left(w\right)={\left(w\right)}^6+4{\left(w\right)}^5+3{\left(w\right)}^4+2{\left(w\right)}^3+\left(w\right)+1$

$f\left(\omega \right)=1+4w^2+3w+2+w+1=4+4w+4w^2=0f\left(w^2\right)={\left(w^2\right)}^6+4{\left(w^2\right)}^5+3{\left(w^2\right)}^4+2{\left(w^2\right)}^3+\left(w^2\right)+1$

$f\left({\omega }^2\right)=1+4w+3w^2+2+w^2+1=4+4w+4w^2=0$
$\because f\left(w\right)=0\&f\left(w^2\right)=0$
Therefore $f\left(x\right)=x^6+4x^5+3x^4+2x^3+x+1$ is divisible by $(x-w)(x-w^2)$.
Ans: D