Question

x+1 is a factor of ${{x}^{n}}+1$ only if
[a] n is an odd integer
[b] n is an even integer
[c] n is a negative integer
[d] n is a positive integer

Answer 1

Hint: Recall factor theorem. According to factor theorem x-a is a factor of p(x) if p(a) = 0. Use factor theorem and hence form an equation in n. Check for which of the options given p(-1) vanishes.

Complete step-by-step answer:
We have $p\left( x \right)={{x}^{n}}+1$
Since x+1 is a factor of p(x), we have
$p\left( x \right)=\left( x+1 \right)g\left( x \right)$, where g(x) is some polynomial in x.
Put x = -1, we get
$p\left( -1 \right)=\left( -1+1 \right)g\left( -1 \right)=0\times g\left( -1 \right)=0$
Hence x = -1 is a zero of the polynomial p(x).
Now, we have $p\left( x \right)={{x}^{n}}+1$
Substituting x = -1 in the expression of p(x), we get
$p\left( -1 \right)={{\left( -1 \right)}^{n}}+1$
We know that ${{\left( -1 \right)}^{x}}=\left\{ \begin{matrix}
   1\text{ if }x\text{ is even} \\
   -1\text{ if }x\text{ is odd} \\
\end{matrix} \right.$
Now if n is odd, we have ${{\left( -1 \right)}^{n}}=-1$
Hence $p\left( -1 \right)=-1+1=0$
If n is even, we have
${{\left( -1 \right)}^{n}}=1$
Hence $p\left( -1 \right)=1+1=2$
Hence if n is an odd integer, then x+1 is a factor of p(x).
Hence option [a] is correct.

Note: [1] We did not check the case when n is negative since in that case ${{x}^{n}}+1$ is not a polynomial at all and hence it being divisible by x+1 is meaningless.