What determines the factors of the multipole expansion? The multiple expansion of a potential V has contributing terms proportional to $\frac{1}{r^{n+1}}$ where $n=0,1,2...$. 
First, why are we interested only in integer powers of r?
Second, why are we interested in symmetry pole arrangements? In other words, the $1/r$ term is a monopole, the $1/r^2$ term is a dipole, the $1/r^3$ term a quadrupole -- all of these are symmetric configurations. Why do we not consider an asymmetric 'tripole' formed in an isosceles triangle? 
The reason I pose these questions is that my understanding of this formalism is that it represents the potential far away from an arbitrarily complicated charge configuration. But if the charge configuration is arbitrarily complicated, then why do we assume such perfectly symmetric poles?
 A: As ACuriousMind pointed out, the integer powers of $1/r$ come from the fact that the multipole expansion is a Laurent expansion, and therefore by definition is an expansion in integer powers of $1/r$.
Regarding your question about symmetry, the point is that each term in the multipole expansion must be "independent" of the others in the sense that it cannot be expanded in terms of them.  For example, any configuration of charges that do not sum to zero will have a monopole contribution, while any configuration is which the positive and negative charges are not symmetric under an appropriate reflection will have a dipole contribution, etc.  Now, your "tripole" configuration can actually be expanded in terms of a monopole term (since any configuration of three equal magnitude charges cannot be electrically neutral) and a dipole term (since the charge distribution of the triangle isn't symmetric under reflections), plus (I believe) other terms.  So in this sense it's not independent of the other multipole moments.
A: 
First, why are we interested only in integer powers of r?

Because when you expand 
$$
\frac{1}{|\vec r - \vec r'|}
$$
about $r'=0$ you get only integer valued terms in the power series expansion.
So, why does this matter? It's because the potential at $\vec r$ due to a localized charge distribution $\rho$ is 
$$
V(\vec r)=\int \frac{\rho(\vec r')}{|\vec r - \vec r'|}d^3r'
=\frac{1}{r}\int \rho (\vec r')d^3r' + \frac{\hat r}{r^2}\cdot \int \rho(\vec r')\vec r'+\mathcal{O}\left(\frac{1}{r^3}\right)
$$
