<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] (This is a comment, not an answer.)
If $f_n(x)$ is your polynomial, starting with $f_0(x)=1$, then
$$ \sum_{n=0}^\infty f_n(x) y^n = 
   \frac{1-xy+x^2y^2+x^2y^3}{(1+xy^2)(1-xy-xy^2)}
   = 1 + \frac{x^2y^2(1+y)^2}{(1+xy^2)(1-xy-xy^2)}. $$
Also, I noticed that $f_n(x)-x f_{n-1}(x)-x f_{n-2}(x)$ only has one or two terms, so a recurrence is possible. That there are no positive real zeros follows from the fact that there are no negative coefficients. The rest of your question is another matter.
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