Fibonacci polynomials and factorization redux I recently asked a question about factorizing a certain expression involving Fibonacci and Lucas polynomials. That question had a very simple and nice answer, but now I've come across another similar question which has me stumped.
Conjecture: The polynomial $4- F_n(x)^4(x^2+3)$ factors as $H_n(x)H_n(-x)$ for some integer polynomial $H_n(x)$, where $F_n(x)$ is the nth Fibonacci polynomial.
Any ideas? 
(In case anybody is wondering why I care about these question, they are related to calculating the Conway polynomial of certain families of knots.)
Edit: As requested, here are the first few examples. Let $G_n(x)=4- F_n(x)^4(x^2+3)$. Then


*

*$G_1(x)=(1+x)(1-x)$

*$G_2(x)=(1+x)(1-x)(2+x^2)^2$

*$G_3(x)=(-1 - 3 x + 2 x^2 - 4 x^3 + x^4 - x^5) (-1 + 3 x + 2 x^2 + 4 x^3 + 
   x^4 + x^5)$

*$G_4(x)= (2 - 4 x + 4 x^2 - 10 x^3 + 4 x^4 - 6 x^5 + x^6 - x^7) (2 + 4 x + 
   4 x^2 + 10 x^3 + 4 x^4 + 6 x^5 + x^6 + x^7)$

*$G_5(x)=(-1 - 5 x + 6 x^2 - 20 x^3 + 11 x^4 - 21 x^5 + 6 x^6 - 8 x^7 + x^8 - 
   x^9) (-1 + 5 x + 6 x^2 + 20 x^3 + 11 x^4 + 21 x^5 + 6 x^6 + 8 x^7 +
    x^8 + x^9)$


Edit 2: Following Will Jagy's ideas, define a family of polynomials $H_n(x)$ by $H_0(x)=2$, $H_1(x)=-1+x$ and then $$H_{n+2}(x)=(x^2+2)H_{n+1}(x)-H_n(x)+(-1)^n(6+2x^2)$$ for $n\geq 0$. Computations show that $$H_n(x)H_n(-x)=G_n(x)$$ for $n\leq 20$. What remains is to prove that this is a valid formula in general.

This looks a little better, $H_1(x) = 1 + x, \; H_2(x) = 2 + 2x + x^2 + x^3, \; H_3(x) = -1 + 3 x + 2 x^2 + 4 x^3 + x^4 + x^5, $ then
$$   H_{n+2}(x) = (2+ x^2) H_{n+1}(x) - H_n(x) + OTHERERROR_{n+2}(x),   $$
where this way the error stays quadratic, by hand calculations the first three error terms come out
$$ -4 + x^2, \; 6 + 2 x^2, \; -6 - 2 x^2... $$
