Question

For all $n\in N$, $x^n+1$ is divisible by

• A. $x+1$
• B. $x-1$
• C. Both (a) and (b)
• D. none of these

Answer 1

The correct option is A $x+1$

Consider the polynomial,

$x^n+1$

Now, we know that,

$x^n+y^n$ is divisible by $x+y$ if $n$ is odd

$\because x^n+y^n=(x+y)(x^{n-1}-x^{n-2}.y+x^{n-3}.y^2-.........y^{n-1})$

$\Rightarrow (x^n+1)=(x^n+1^n)$

$\Rightarrow (x^n+1^n)=(x+1)(x^{n-1}-x^{n-2}.1+x^{n-3}{.1}^2-{.........1}^{n-1})$

Hence, $(x^n+1)$is divisible by $(x+1)$ if $n$ is odd

Hence,this is the answer.