<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] Hint: If $x_0=-1$ is a zero of $x^{2016}-1$, then $(x-(-1))$ is a factor of $(x^{2016}-1)$.
In order to find $G(x)=(x^{2016}-1)/(x+1)$ you need to apply polynomial long division (you should see the pattern).
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