Question

If A(x) and B(x) to the polynomials and f$\left(x\right)=A\left(x^3\right)+xB\left(x^3\right)$. If f(x) is divided by $x^2+x+x+1$ , then it is divisible by x- 1 also

Answer 1

f$\left(x\right)=A\left(x^3\right)+xB\left(x^3\right)+1=0$
$\Rightarrow (x-\omega )(x-{\omega }^2)=0$ By given the condition $\omega ,{\omega }^2$ will also the root of $f\left(x\right)=0$ and ${\omega }^n=1$
$\therefore f\left(\omega \right)=0\Rightarrow A\left(1\right)+\omega B\left(1\right)=0$

$f\left({\omega }^2\right)=0\Rightarrow A\left(1\right)+{\omega }^{2}B\left(1\right)=0$
sloving (1) and (2) we get $A\left(1\right)=0,B\left(1\right)=0$ showing that both $A\left(x\right)andB\left(x\right)$ are divided for $x-1$
$\therefore f\left(x\right)=A\left(x^3\right)+xB\left(x^3\right)+1=0$