AMC10 2025 B

Observe that both $f(n)$ and $g(n)$ can be factored using RZT. By trying out a few values, we get that $n = -1$ is a root of $f(n)$. Then, we can use synthetic division (regular polynomial division works, too) to get the other roots of $f(n)$, which are $n = 4$ and $n = 2$. Now we factor $g(n)$. By inspection, $n = -1$ also works on $g(n)$, and so synthetic division and factoring the remaining quadratic gets us the roots $n = 3$ and $n = 4.$ We can now express $\frac{f(n)}{g(n)}$ as: \[\frac{(n + 1)(n - 4)(n - 2)}{(n + 1)(n - 4)(n - 3)}.\] We can cancel the top and bottom to get: \[\frac{(n - 2)}{(n - 3)}\] Notice that the only values for which this expression is an integer are $n = 4$ and $n = 2$, because they both cause the denominator to have a magnitude of $1$. However, $n = 4$ is an extraneous solution as plugging it back into the original expression would make it evaluate to $\frac{0}{0}$. Our answer is then $\boxed{\textbf{(A)}\hspace{3pt} 2}$.