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.
 A: 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 $n$.
A: EDITED. First let us define
$$g_n(x) := \sum_{k=0}^{\lfloor n/2\rfloor} {n-k\choose k} x^{n-k}.$$
Then empirically,
$$\eqalign{f_{2n}(x) &= g_n(x)^2\cr
   f_{4n+1}(x) &= g_{2n+2}(x)g_{2n-1}(x)\cr
   f_{4n+3}(x) &= g_{2n}(x)g_{n+1}(x)h_{n+2}(x)\cr}$$
where $h_n(x)$ is a Lucas polynomial.
I think it should be possible to prove these formulas using the recurrence $f_n(x)=xf_{n−1}(x)+x^2f_{n−3}(x)+x^2f_{n−4}(x)$. Then, as noted in Ira Gessel's comment below, the desired properties of the roots of $f_{2n}(x)$ and $f_{4n+1}(x)$ follow from the properties of the Chebyshev polynomials. Presumably the roots of Lucas polynomials are also well understood, but I'm not so familiar with them.
A: (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.
A: 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.
A: (another comment, not an answer.)
Experimentally, with the following code
from sympy import *
var('x')
var('n', integer = True)

f = ( (2+(x+3) * sqrt(-x))/(2*(x+4)) *
      sqrt(-x)**n + ((2-(x+3) * sqrt(-x)) / (2*(x+4))) * (-sqrt(-x))**n +
      + (x+2+sqrt(x*(x+4)))/(2*(x+4)) *
        ( (x+sqrt(x*(x+4)))/2 )**n +
        (x+2-sqrt(x*(x+4)))/(2*(x+4)) *
        ( (x-sqrt(x*(x+4)))/2 )**n
        )
pprint(f)
for i in range(10):
   eq = simplify(f.subs(n,i))
   print ('========== n = ', i)
   pprint(eq)
   print ('have solutions')
   sols = solve(eq)
   pprint (sols)
   pprint ('approx. = ')
   pprint ([s.evalf() for s in sols])

we see the following:
$$
\begin{array}{l}
f_{0} = 1\\
f_{1} = 0\\
f_{2} = x^{2}\\
f_{3} = x^{2} \left(x + 2\right)\\
f_{4} = x^{2} \left(x^{2} + 2 x + 1\right)\\
f_{5} = x^{3} \left(x^{2} + 3 x + 1\right)\\
f_{6} = x^{4} \left(x^{2} + 4 x + 4\right)\\
f_{7} = x^{4} \left(x^{3} + 5 x^{2} + 7 x + 3\right)\\
f_{8} = x^{4} \left(x^{4} + 6 x^{3} + 11 x^{2} + 6 x + 1\right)\\
f_{9} = x^{5} \left(x^{4} + 7 x^{3} + 16 x^{2} + 13 x + 2\right)\\
\end{array}
$$

And it seems that already $f_9$ have some imaginary roots, albeit very small.
Do these imaginary roots really converge to 0 when $n\to \infty$ or they stay on the same magnitude ?
A: Take the first two terms in your sum, and multiply by $x+4$, then
you get a sequence of polynomials which satisfy $P_n = -x P_{n-2}$,
and if you do the same for the last two terms, you get a sequence
determined by $Q_n = x Q_{n-1}+x Q_{n-2}$.
Now, $Q_n$ interlaces the roots of $Q_{n+1}$ so these are easy to show
are real-rooted. Same goes for $P_n$ and $P_{n+2}$.
So, perhaps one can combine these somehow in order to show that the sum is real-rooted.
