Question

Let $P\left(x\right)$ and $Q\left(x\right)$ be two polynomials. Suppose that $f\left(x\right)=P\left(x^3\right)+xQ\left(x^3\right)$ is divisible by $x^2+x+1$, then

• A. $P\left(x\right)$ is divisible by $(x-1)$, but $Q\left(x\right)$ is not divisible by $(x-1)$
• B. $Q\left(x\right)$ is divisible by $(x-1)$, but $P\left(x\right)$ is not divisible by $(x-1)$
• C. Both $P\left(x\right)$ and $Q\left(x\right)$ are divisible by $(x-1)$
• D. $f\left(x\right)$ is divisible by $(x^3-1)$

Answer 1

Answer: (C), (D)

The correct options are
C Both $P\left(x\right)$ and $Q\left(x\right)$ are divisible by $(x-1)$
D $f\left(x\right)$ is divisible by $(x^3-1)$

$x^2+x+1=(x-w)(x-w^2)$, where $w$ is cube root of unity.
Since, $f\left(x\right)$ is divisible by $x^2+x+1$.
$\therefore f\left(w\right)=0,f\left(w^2\right)=0$
So, $P\left(w^3\right)+wQ\left(w^3\right)=0$
$\Rightarrow P\left(1\right)+wQ\left(1\right)=0...\left(1\right)$
$P\left(w^6\right)+w^{2}Q\left(w^6\right)=0$
$\Rightarrow P\left(1\right)+w^{2}Q\left(1\right)=0...\left(2\right)$
Solving eqn$\left(1\right)$ and $\left(2\right)$, we get
$P\left(1\right)=0$ and $Q\left(1\right)=0$