Polynomial division first!

Question

$$ x^{28}+9x^{26}+39x^{24}+95x^{22}+76x^{20}-384x^{18}-1928x^{16}-4868x^{14}-7712x^{12}-6144x^{10}+4864x^{8}+24320x^{6}+39936x^{4}+36864x^{2}+16384 $$

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?

Answer 1

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 $x\mapsto 2/x$.

Answer 2

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.

Answer 3

Let $P_{28}(x)$ is the issue polynomial and $$R(x) = x^{-14}P_{28}(x).$$

Then $$C(t) = R(e^{it}\sqrt2) = 4(64\cos14t+288\cos12t+624\cos10t$$ $$+760\cos8t+304\cos6t-768\cos4t-1928\cos2t-1217),$$ and $$P_7(y) = C\left(\dfrac12\arccos\dfrac y4\right) = y^7+9y^6+11y^5-121y^4-480y^3-608y^2-168y+92.$$

Another way is using of the Chebyshev Polynomials of the First Kind, then $$P_7(y) = 4(64T_7\left(\dfrac y4\right) + 288T_6\left(\dfrac y4\right) + 624T_5\left(\dfrac y4\right) + 760T_4\left(\dfrac y4\right)$$ $$ + 304T_3\left(\dfrac y4\right) - 768T_2\left(\dfrac y4\right) - 1928 \dfrac y4 - 1217),$$ with the same result (idea of OP author, see the comments).

Easily to show that $P_7(y)$ has only one root out of the interval $(-4,4)$ (proposition of the OP author, see the comments).

Since $$P'_7(t+4) = 7t^6 + 222t^5 + 2815t^4 + 17996t^3 + 59472t^2 + 90240t + 39000$$ and $$P_7(4) = -29968 <0,$$ then $P_7(y)$ has a single root on the interval $(4,\infty).$

On the other hand, $$P'_7(-4-t) = 7 t^6 + 114 t^5 + 655 t^4 + 1684 t^3 + 1968 t^2 + 896 t + 88$$ and $$P_7(-4) = 16 >0,$$ so there are not roots in the interval $(-\infty,-4).$

At the same time, the plot of $P_7(7)$

shows that there are six different roots in the interval $(-4,1),$ which can be easily separated.

Approximately, the roots are $$y_k\in\{4.16023,0.25701,-1.08886,-2.29038,-2.31414,-3.78887,-3.93500\}.\tag1$$

Since $$y=4\cos2t,\quad t = -i\ln\dfrac x{\sqrt2},$$ then $$y=4\cos\left(-2i\ln\dfrac x{\sqrt2}\right) = 2\left(e^{\large2\ln\frac x{\sqrt2}} + e^{-\large2\ln\frac x{\sqrt2}}\right) = x^2 + \dfrac4{x^2},$$ $$y-y_k = x^2-4\dfrac {y_k}4 + \dfrac4{x^2},$$ $$P_{28}(x) = \prod\limits_{k=0}^6\left(x^4-4\dfrac{y_k}4 x^2+4\right),\tag2$$

wherein $$y_0 > 4,\quad |y_{1-6}| <4.$$ Therefore, required decomposition of $P_{28}(x)$ is confirmed.