A question on the real root of a polynomial For $n\geq 1$, given a polynomial
\begin{equation*}
  \begin{aligned}
   f(x)=&\frac{2+(x+3)\sqrt{-x}}{2(x+4)}(\sqrt{-x})^n+\frac{2-(x+3)\sqrt{-x}}{2(x+4)}(-\sqrt{-x})^n \\
   &+\frac{x+2+\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x+\sqrt{x(x+4)}}{2} \right )^n+\frac{x+2-\sqrt{x(x+4)}}{2(x+4)}\left ( \frac{x-\sqrt{x(x+4)}}{2} \right )^n.
  \end{aligned}
  \end{equation*}
Using Mathematic $12.3$, when $n$ is large enough, we give the distribution of the roots of $f(x)$ in the complex plane as follows

In this figure, we can see that the closure of the real roots of $f(x)$ may be $\left [ -4,0 \right ]$.
So we have the following question
Question: all roots of $f(x)$ are real? It seems yes! But we have no way of proving it.

Assume that we have $x = - (t + 1/t)^2 = -t^2 - 1/t^2 - 2$ for some $t \in \mathbb{C}$. Then
$$
\sqrt{-x} = t + \frac{1}{t} = \frac{t^2 + 1}{t},\quad x + 4 = -\left(t - \frac{1}{t}\right)^2, \quad \sqrt{x(x + 4)} = t^2 - \frac{1}{t^2},
\\
x + \sqrt{x(x + 4)} = -2\left(1 + \frac{1}{t^2}\right) = -2\frac{1 + t^2}{t^2}, \quad x - \sqrt{x(x + 4)} = -2(1 + t^2)
$$
Consequently, we get
$$
f_n(x)= \tilde f_n(t) = \frac{(1 + t^2)^n(t^{n + 2} - 1)^2}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is even},
\\
f_n(x) = \tilde f_n(t) = -\frac{(1 + t^2)^n(1 - t^{n -1})(1 - t^{n+5})}{t^{2n}(t^2 - 1)^2},\quad\text{if } n\text{ is odd}.
$$
Hence the only roots of $\tilde f_n$ are the suitable roots of unity and the roots of $f_n$ are all in the real segment $[-4, 0]$ and their closure is the whole segment for large $n$.
