Let me recall the definition which seems the most standard of Fricke polynomials. 
Let $G$ be the free group with two generators $u,v$. It is not very hard to prove that there exists a unique application $t : G \rightarrow \mathbb{Z}[X,Y,Z]$
such that 


(1) $t(u)=X,t(v)=Y,t(uv)=Z$,


(2) $t$ is the character (i.e. the trace) of a representation $r : G \rightarrow \rm{SL}\_2(K)$ where $K$ is some field containing $\mathbb{Z}[X,Y,Z]$.


A *Fricke polynomial* is just an element of $\mathbb{Z}[X,Y,Z]$ which is in the mage of $t$.



> 
> What is the image of $t$ ? 
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Edit: As requested, I give here the proof of the statement about $t$. I believe that this statement is due in some form to Fricke in the nineteenth century: 


If $t:G \rightarrow K$ is
the trace of a representation $r: G \rightarrow {\rm{SL}}\_2(K)$, then one has $t(gh)+t(gh^{-1}) = t(g)t(h)$ for all $g,h$ in $G$ (since this formula holds when $t$ is the trace, and $g,h$ are matrices in ${\rm SL}\_2(K)$, as easily checked). Now this formula, and an easy induction on the length
of a word $g$ in $G$ shows that $t(g)$ can be expressed as a polynomial (with integral coefficients) in $t(u)$, $t(v)$ and $t(uv)$. Hence a $t$ satisfying (1) and (2) is unique if it exists. 


For the existence, let $U$ and $V$ be two matrices in ${\rm SL}\_2(\mathbb{C})$ such that tr$U$,
tr$V$, and tr$UV$ are algebraically independent (this is really easy). Then see $\mathbb{Z}[X,Y,Z]$ as a subring of $\mathbb{C}$ by sending $X$ on tr$U$, $Y$ on tr$V$, $Z$ on tr$UV$.
Let $r$ be the representation $G \rightarrow \rm{SL}\_2(\mathbb{C})$ sending $u$ on $U$ and $v$ on $V$, and $t=$tr $r$. Then by definition $t$ satisfies (2), with $K=\mathbb{C}$, one has $t(u)=X$, $t(v)=Y$, $t(uv)=Z$ by construction and $t$ takes value in the subring $\mathbb{Z}[X,Y,Z]$ of $\mathbb{C}$ by what we have said earlier. Hence the existence of $t$.



Which polynomials are Fricke polynomials ?
I do not think that there is a complete answer to this question. However, one can give some necessary conditions (which show that any answer must be complicated).


One can show that a triple $(x,y,z) \in \mathbb C^3$ comes from traces of matrices in $SU(2)$ (i.e. there exist $u,v \in SU(2)$ such that $(x,y,z)=(t(u),t(v),t(uv))$) if and only if $(x,y,z) \in \mathbb [-2,2]^3$ (this is obviously necessary) and $$x^2 + y^2 + z^2 - xyz \leq 4.$$


Hence, a necessary condition on a polynomial $p \in \mathbb Z[X,Y,Z]$ to be a Fricke polynomial is that for all $(x,y,z) \in [-2,2]^3$ we have the implication
$$x^2 + y^2 + z^2 - xyz \leq 4 \quad \Rightarrow \quad p(x,y,z) \in [-2,2].$$


Indeed, the word evaluated on $u$ and $v$ will give again a matrix in $SU(2)$ and hence its trace is in the interval $[-2,2]$.


The semi-algebraic set $$S:=[-2,2]^3 \cap \lbrace (x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 - xyz \leq 4 \rbrace$$ is a spectrahedron (that means it can be defined by a linear matrix inequality) and is called the Elliptope $E\_3$.


