How do I resolve a recurrence relation when the characteristic equation has fewer roots than terms?

The characteristic equation is actually $x^3-9x^2+15x+25 = 0$; it doesn’t depend on $n$. After factoring this becomes $(x+1)(x-5)^2 = 0$, with a double root at $x=5$. In this case the general solution has the form $a_n = \alpha_1(-1)^n + \alpha_2 \cdot 5^n + \alpha_3n \cdot 5^n$, and you can use the known values of $a_0,a_1,a_2$ to solve for $\alpha_1,\alpha_2,\alpha_3$.

More generally, if $r$ is a root of the characteristic equation of multiplicity $m$, it gives rise to these $m$ terms in the general solution:$$\alpha_1r^n + \alpha_2nr^n + \alpha_3n^2 r^n + \dots + \alpha_m n^{m-1}r^n.$$Thus, you will always have as many terms as the degree of the characteristic equation.

Factor the characteristic polynomial to get $$ x^3-9x^2+15x+25=(x+1)(x-5)^2 $$ The $x+1$ factor requires a term of the form $a(-1)^k$, but the $(x-5)^2$ term requires $(b+ck)5^k$. This is because both $5^k$ and $k\:5^k$ are annihilated by the difference operator $(S-5)^2$ (where $S$ is the shift operator: $Sa_n=a_{n+1}$). Now, just find $a$, $b$, and $c$ to fit your initial data.

For factors of $(x-a)^n$, use $(b_0+b_1 k+b_2 k^2+...+b_{n-1}k^{n-1})a^k$ since this is annihilated by $(S-a)^n$.