<s>[INST] Is $x+1$ a factor of $x^{2016}-1$? $$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$
If $x+1$ is a factor of $x^{2016}-1$, then $(1+x+x^2+x^3+\dots+x^{2015})=(x+1)G(x)$, where $G(x)$ is some polynomial.
What is $G(x)$ if $x+1$ is also a factor?
 [/INST] Yes, $x+1$ is a factor of $x^{2016}-1$ and
$$x^{2016}-1=(x+1)(x^{2015}-x^{2014}+x^{2013}-\cdots-x^2+x-1)$$
Then
$$G(x)=1+x^2+x^4+\dots+x^{2014}$$
More generally, $x-r$ is a factor of $P(x)$ if and only if $P(r)=0$. In this case, $P(x)=x^{2016}-1$ and $P(-1)=0$, so $x+1$ is a factor of $x^{2016}-1$.
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