Is $x+1$ a factor of $x^{2016}-1$?

Question

$$x^{2016}-1=(x-1)(1+x+x^2+x^3+\dots+x^{2015})$$ 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?

Answer 1

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).

Answer 2

Use this simple lemma

$(x-a)$ is a factor of $P(x)$ if and only if $P(a)=0$

Answer 3

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 $x^{2016}-1$.

Answer 4

\begin{align} x^{2016}-1=&(x^2)^{1008}-1=(x^2-1)((x^2)^{1007}+\dots+x^2+1)\\ &=(x+1)(x-1)(x^{2014}+\dots+x^2+1). \end{align}