Polynomial division first! $$
x^{28}+9x^{26}+39x^{24}+95x^{22}+76x^{20}-384x^{18}-1928x^{16}-4868x^{14}-7712x^{12}-6144x^{10}+4864x^{8}+24320x^{6}+39936x^{4}+36864x^{2}+16384
$$
In an example the polynomial above arose. It should have six factors, each like 
$$
(x^4-4x^2\cos(4\pi c_\color{blue}{\text{r}}) +4)
$$
and one like $$
(x^4-4x^2\cosh(4\pi c_\color{red}{\text{n}})+4).
$$
to match the degree $6\cdot4+4=28$ of the first.
Is it possible to first do the polynomial division and then determine the values of $\cos(4\pi c)$?
Or the other way: Why not just multiplying all the factors to get a set of equation? These are elementary symmetric polynomials in the variables $\cos(h)(4\pi c_{r/n})$ with integer solutions. 

Is this approach known to work for a class of cases? What can we say about the angles $\cos(h)(4\pi c_{r/n})$ because of that elementary symmetric polynomial integer solution contraint?

 A: Let $f(x)$ be your polynomial. It has a number of symmetries. Dietrich Burde's answer exhibits one. We see another by calculating
$$g(u):=2^{-9}f(\sqrt{2u})=32 u^{14}+144 u^{13}+312 u^{12}+380 u^{11}+152 u^{10}-384 u^9-964 u^8-1217 u^7-964 u^6-384
   u^5+152 u^4+380 u^3+312 u^2+144 u+32.$$
A most obvious feature of this is that it is palindromic. Meaning that $u=\xi$ is a zero of $g(u)$ if and only if $u=1/\xi$ is. A common trick taking advantage of this is to write everything in terms of the new variable $v=u+1/u$. It is simple to verify that
$$
u^{-7}g(u)=P(v),
$$
with
$$
P(v)=32 v^7+144 v^6+88 v^5-484 v^4-960 v^3-608 v^2-84 v+23.
$$
Mathematica thinks that $P(v)$ is irreducible over $\Bbb{Q}$. Judging from the plot it has five zeros in the interval $(-2,0)$ and two positive zeros approximately $0.1$ and $2.1$. If $u$ is real, then $v=u+1/u$ has absolute value $\ge2$. This would yield four real zeros of $f(x)$ subject to symmetries: $x\mapsto -x$ and $x\mapsto 2/x$.
A: The polynomial has two irreducible factors over $\Bbb Q$, namely $f=pq$ with
$$
p=x^{14} + 3x^{13} + 9x^{12} + 21x^{11} + 42x^{10} + 78x^{9} + 124x^{8} + 190x^{7} + 248x^{6} +
312x^{5} + 336x^{4} + 336x^{3} + 288x^{2} + 192x + 128,
$$
and
$$
q=
x^{14} - 3x^{13} + 9x^{12} - 21x^{11}
+ 42x^{10} - 78x^9 + 124x^8 - 190x^7 + 248x^6 - 312x^5 + 336x^4 - 336x^3 + 288x
^2 - 192x + 128.
$$
Note that the sign alternates for $q$. Now it is a bit easier to come to the six factors you want.
