$(x + 1)$ is a factor of $x^{n} + 1$ when

$(x + 1)$ is a factor of $x^{n} + 1$ when
  1. $n$ is odd
  2. $n$ is even
  3. $n = 0$
  4. never

Solution

$(-1)^{n} + 1 = 0 \Rightarrow (-1)^{n} = -1 \Rightarrow n$ odd.

Asked in: MH-SSC-9

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