<s>[INST] 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.
 [/INST] Here is a proof, using Maple calculations, of Tim Chow's empirical observations. We use Hadamard products of power series.  The Hadamard product (with respect to the variable $y$) is defined by
$$
\sum_{n=0}^\infty a_n y^n *\sum_{n=0}^\infty b_n y^n= \sum_{n=0}^\infty a_n b_n y^n.
$$
The Hadamard product of two rational power series is rational, and I did the following computations with a Maple program I wrote using the method described here .
For any power series $A(y) = \sum_{n=0}^\infty a_n y^n$ and  integers $m$ and $i$, let
\begin{equation*}
A_{m,i}(y) = \sum_{n=0}^\infty a_{mn+i}y^i,
\end{equation*}
where we take $a_n=0$ if $n<0$. Following Brendan McKay, we  define the generating function.
$$F=F(y) = \sum_{n=0}^\infty f_n(x) y^n = 
  1+\frac{x^{2} y^{2} \left(1+y \right)^{2}}{\left(1+x y^{2}\right) \left(1-xy-x y^{2}\right)}
$$
We also define generating functions for Timothy Chow's polynomials $g_n(x)$ and $h_n(x)$:
\begin{gather*}
G=\sum_{n=0}^\infty g_n(x) y^n = \frac{1}{1-xy-xy^2}\\
H=\sum_{n=0}^\infty h_n(x) y^n = \frac{2-xy}{1-xy-xy^2}.
\end{gather*}
Then we want to prove
\begin{gather}
F_{2,0}=G*G\tag{1}\\
F_{4,1}=G_{2,2}*G_{2,-1}\tag{2}\\
F_{4,3}=G_{2,0}*G_{1,1}*H_{1,2}\tag{3}
\end{gather}
Multisections, can be computed by Hadamard products (or in other ways). For example,
$F(y)*y/(1-y^4) = yF_{4,1}(y^4)$. We find that
\begin{gather*}
F_{2,0} =\frac{1-x y }{\left(1+x y \right) \left(1-2xy -x^{2} y +x^{2} y^{2}\right)}\\
F_{4,1}=\frac{x^3y(1+3x+x^2 -x^2y)}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\
F_{4,3}=\frac{x^2(2+x -\left(3x^2+4x^3+x^4\right) y +x^{4} y^{2})}{(1-x^2y)(1-(2x^2+4x^3+x^4) y +x^{4} y^{2})}\\
G_{1,1}=\frac{x(1+y)}{1-xy-xy^2}\\
G_{2,-1}=\frac{xy}{1-2xy-x^2y+x^2y^2}\\
G_{2,0}=\frac{1-xy}{1-2xy-x^2y+x^2y^2}\\
G_{2,2}=\frac{x(1+x-xy)}{1-2xy-x^2y+x^2y^2}\\
H_{1,2}=\frac{x(2+x+xy)}{1-xy-xy^2}
\end{gather*}
We can then verify $(1)$–$(3)$ directly.
</s>