Question

If $f\left(x\right)$ and $g\left(x\right)$ are two polynomials such that the polynomial $P\left(x\right)=f\left(x^3\right)+xg\left(x^3\right)$ is divisible by $x^2+x+1$, then $P\left(1\right)$ is equal to:

Answer 1

Step 1: Apply Factor Theorem.

We know that $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$. So, we can write

$x^2+x+1=\left(x-\omega \right)\left(x-{\omega }^2\right)$.

$\Rightarrow \omega and{\omega }^2$ are the factors of $P\left(x\right)$, so $P\left(\omega \right)=0$ and $P\left({\omega }^2\right)=0$.

So,

.$\begin{array}{rcl}P\left(\omega \right)& =& f\left({\omega }^3\right)+\omega g\left({\omega }^3\right)\\ \Rightarrow 0& =& f\left(1\right)+\omega g\left(1\right)...\left(1\right)\end{array}$

and

$\begin{array}{rcl}P\left({\omega }^2\right)& =& f{\left({\omega }^3\right)}^2+{\omega }^{2}g{\left({\omega }^3\right)}^2\\ \Rightarrow 0& =& f\left(1\right)+{\omega }^{2}g\left(1\right)...\left(2\right)\end{array}$

Step 2: Find the value of $P\left(1\right)$.

Subtracting $\left(2\right)$ from $\left(1\right)$, we get

$\left(\omega -{\omega }^2\right)g\left(1\right)=0$

Now, $\omega -{\omega }^2\ne 0$, so $g\left(1\right)=0$.

By substituting the value of $g\left(1\right)$in $\left(1\right)$, we get

$f\left(1\right)=0$

So,

$\begin{array}{rcl}P\left(1\right)& =& f\left(1^3\right)+1g\left(1^3\right)\\ & =& f\left(1\right)+1g\left(1\right)\\ & =& 0\end{array}$

Therefore, the value of $P\left(1\right)$ is $0$.