associated to the second order recurrence relation $G_{n}(x)$. That is, $a(x)$ and $b(x)$ are the solutions of $z^2-d(x)z-g(x)=0$. The explicit formula for $G_{n}(x)$ given in (\[solutionrecurrencerelationuno\]) with $G_{0}(x)=p_{0}(x)$ and $G_{1}(x)=p_{1}(x)$ imply that $$\label{solutionrecurrencerelationdos}
 t_{1}=\dfrac{p_{1}(x)-p_{0}(x)b(x)}{a(x)-b(x)} \text{ and } t_{2}=\dfrac{-p_{1}(x)+p_{0}(x)a(x)}{a(x)-b(x)}$$

Using (\[solutionrecurrencerelationuno\]) and (\[solutionrecurrencerelationdos\]) we obtain the Binet formulas for the Generalized Fibonacci sequences of the Lucas type and Fibonacci type. Thus, substituting $2p_{1}(x)=p_{0}(x)d(x)$ in (\[solutionrecurrencerelationdos\]) we obtain that $t_{1}=t_{2}= p_{0}(x)/2$. Substituting this in (\[solutionrecurrencerelationuno\]) and letting $\alpha$ be $2/p_{0}(x)$ we obtain the Binet formula for Generalized Fibonacci sequence of *Lucas type* or *first type* $$\label{bineformulados}
L_n(x)=\frac{a^{n}(x)+b^{n}(x)}{\alpha}.$$ We want $\alpha$ be an integer, therefore $|p_{0}(x)|=1 \text { or } 2$.

Now, substituting $p_{0}(x)=0$ and the constant $p_{1}(x)$ in (\[solutionrecurrencerelationdos\]) we obtain that $t_{1}=t_{2}=p_{1}(x)$. Substituting this in (\[solutionrecurrencerelationuno\]) we obtain the Binet formula for Generalized Fibonacci sequence of *Fibonacci type* or *second type* $$\label{bineformulaunogeneral}
R_n(x)=\frac{p_{1}(x)\left(a^{n}(x)-b^{n}(x)\right)}{a(x)-b(x)}.$$ In this paper we are interested only on $R_n(x)$ when $p_1(x)=1$. Therefore, the Binet formula $R_n(x)$ that we use here is $$\label{bineformulauno}
R_n(x)=\frac{a^{n}(x)-b^{n}(x)}{a(x)-b(x)}.$$

Note that $a(x)+b(x)=d(x)$, $a(x)b(x)= -g(x)$ and $a(x)-b(x)=\sqrt{d^2(x)+4g(x)}$ where $d(x)$ and $g(x)$ are the polynomials defined on the generalized Fibonacci polynomials.

A generalized Fibonacci polynomial which satisfies the Binet formula (\[bineformulados\]) is said to be of 