and $z_2(x)=\gcd(q(x),s(x))$, then $$\gcd(p(x)q(x),r(x)s(x))=\frac{\gcd( z_2(x)p(x),z_1(x)s(x))\gcd( z_1(x)q(x), z_2(x)r(x))}{z_1(x)z_2(x)}.$$

We prove part (3). Since $\gcd(p(x),r(x))=z_1(x) $ and $\gcd(q(x),s(x))=z_2(x)$, there are polynomials $P(x)$, $S(x)$, $Q(x)$ and $R(x)$ with $\gcd(P(x),R(x))=\gcd(Q(x),S(x))=1$, such that $ p(x) = z_1(x)  P(x),\;   s(x)= z_2(x)  S(x),\;  r(x) = z_1(x)  R(x),   \;
q(x)= z_2(x)  Q(x)$. So, $$\begin{aligned}
\gcd(p(x)q(x),r(x)s(x)) &=&\gcd(z_1(x) P(x) z_2(x)Q(x), z_1(x) R(x) z_2(x)S(x))\\
&=&z_1(x)z_2(x)\gcd( P(x)Q(x), R(x)S(x)).\\\end{aligned}$$ From part (2) we know that $
\gcd( P(x)Q(x), R(x) S(x))=\gcd( P(x),S(x))\gcd( Q(x), R(x)).
$ Now it is easy to see that $$\gcd(p(x)q(x),r(x)s(x))=\frac{\gcd( z_2(x)p(x),z_1(x)s(x))\gcd( z_1(x)q(x), z_2(x)r(x))}{z_1(x)z_2(x)}.$$ This proves part (3).

We recall that $\rho=\gcd(d(x),G_1(x))$ and that for GFP of Lucas type it is required that $\gcd(p_0(x), p_1(x))=1$, $\gcd(p_0(x), d(x))=1$, $\gcd(p_0(x), g(x))=1$ and $\gcd(d(x), g(x))=1$. We also recall $p_0(x)=0$ and $p_1(x)=1$ for GFP of Fibonacci type.

For the rest of the paper we use the notation $G_n^{*}$ if the generalized Fibonacci polynomial $G_n$ satisfies the Binet formula (\[bineformulados\]) and $G_n^{'}$ if the generalized Fibonacci polynomial $G_n$ satisfies the Binet formula (\[bineformulauno\]). We use $G_n$ if the result does not need the mentioned classification to be true. We recall that for Lucas type polynomials $|p_{0}(x)| =1 \text{ or } 2$ and for Fibonacci type polynomial $p_{1}(x) =1$.

\[gcdlemmas\] If $\{ G_{n}(x)\}$ is a GFP of either 