How can I find a projective resolution of the $k[x]$-module $k[x]\oplus \frac{k[x]}{(x^m)}$ (where $k$ is a field and $m\in\mathbb{Z}$)? That is, I want to find a long exact sequence $$\cdots\rightarrow P_{n+1}\rightarrow P_n\rightarrow\cdots\rightarrow P_0\rightarrow k[x]\oplus \frac{k[x]}{(x^m)}\rightarrow 0$$ where $P_i$ are all projective $k[x]$-modules. This is part of a larger exercise, but I can't get started without the projective resolution.
Edit: For my purposes, I would also be happy with a free resolution of $$k[x]\oplus \frac{k[x]}{(x^m)}$$
Hint:
Take the direct sum of projective resolutions of each summand. The first summand is free, so its projective resolution is rather simple. The second summand is the cokernel of multiplication by $x^m$ in $k[x]$. So you get a finite free resolution of length $1$ (which is normal since a P.I.D. has global dimension $1$).