Question

Show that $p-1$ is a factor of $p^{10}-1$ and also of $p^{11}-1$.

Answer 1

Step-1: Calculating the value of $p$:

Factor Theorem:-$p\left(x\right)$ be a polynomial of degree greater than or equal to 1 and a be a real number. $p\left(a\right)=0$, then $(x-a)$ is a factor of $p\left(x\right)$.

Given equations, $p^{10}-1$and $p^{11}-1$

From the Factor's theorem,

Let, $p-1=0$

$p=1$

Step-2: Finding the value of both the polynomials at $p=1$:

For $p-1$ to be a factor, the equation on substituting $p=1$should equate to $0$

$$\begin{gathered} \left(i\right)f\left(p\right)=p^{10}-1 \\ \Rightarrow f\left(1\right)=1-1=0 \end{gathered}$$

$$\begin{gathered} \left(ii\right)f\left(p\right)=p^{11}-1 \\ \Rightarrow f\left(1\right)=1-1=0 \end{gathered}$$

The values of both the polynomials is $0$ at $p=1$.

Hence proved, $p-1$ is a factor of both $p^{10}-1$and $p^{11}-1$.