Factor $1+x+x^2+x^3+...+x^{14}$ In a previous task, I was asked to factor $1+x+x^2+x^3$ for $x \in \mathbb{R}$, which I accomplished by solving 
$1+x+x^2+x^3 = 0 \to $
$1+x(1+x+x^2) = 0 \to $
$x(1+x+x^2) = -1$
which has a solution $x = -1$, and thus I knew $(x+1)$ was a factor. A bit of guesswork gave me $(x+1)(x^2+1)$.
Now I'm asked to factor $1+x+x^2+x^3+...+x^{14}$ for $x \in \mathbb{R}$ and I'm a bit stuck. Again, we have the implication $x(1+x+x^2+...+x^{13}) = -1$, for which $x=-1$ is a solution, so again we have a factor $x+1$. But now I cannot apply guesswork to determining the rest of the factors, so I feel there is some kind of conclusion I can draw about the powers (perhaps their parity) to solve this problem?

Using the fact
$$ x^n-1 = (x-1)(1+x+x^2+\dots+x^{n-1}),$$
our polynomial can be written in the form
$$ 1+x+x^2+x^3+…+x^{14} = \frac{x^{15}-1}{x-1}. $$
Now, we can find the roots of $ x^{15} - 1 $ using the complex variable tecniques
$$ x^{15}=1=e^{i2k\pi} \implies x = e^{\frac{i2k\pi}{15}},\quad k=0,1,2,\dots,14. $$
So, our polynomial can be written as
$$ 1+x+x^2+\dots+x^{14} =(x-e^{\frac{i2\pi}{15}})(x-e^{\frac{i4\pi}{15}})\dots (x-e^{\frac{i28\pi}{15}})$$ 
$$ = \Pi_{m=1}^{14}(x-e^{\frac{i2m\pi}{15}}). $$ 
Note that,
$$ e^{i\theta}= cos(\theta)+i\sin(\theta) $$ 
$$  e^{2k\pi i} = 1,\quad k\in \mathbb{Z}. $$
