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 $x-1$

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 $x-1$

We have,
$x^2+x+1=(x-\omega )(x-{\omega }^2)$
Since $f\left(x\right)$ is divisible by $x^2+x+1,f\left(\omega \right)=0,f\left({\omega }^2\right)=0$, so
$P\left({\omega }^3\right)+\omega Q\left({\omega }^3\right)=0\Rightarrow P\left(1\right)+\omega Q\left(1\right)=0$ (1)
$P\left({\omega }^6\right)+{\omega }^{2}Q\left({\omega }^6\right)=0\Rightarrow P\left(1\right)+{\omega }^{2}Q\left(1\right)=0$ (2)
Solving (1) and (2), we obtain
$P\left(1\right)=0$ and $Q\left(1\right)=0$
Therefore, both $P\left(x\right)$ and $Q\left(x\right)$ are divisible by $x-1$. Hence, $P\left(x^3\right)$ and $Q\left(x^3\right)$ are divisible by $x^3-1$ and so by $x-1$ Since $f\left(x\right)=p\left(x^3\right)+xQ\left(x^3\right)$, we get $f\left(x\right)$ is divisible by $x-1$.