I assume that you mean
$$
f(x)=\sum_{i=0}^nc_ix^i
$$
In that case, we are looking at
$$
f(x)=R(x)-2R(x-1)-3R(x-2)=(I+S)(I-3S)R(x)\tag{1}
$$
where $S$ is the shift operator $Sf(x)=f(x-1)$. Since the general solution of
$$
R_0(n)-2R_0(n-1)-3R_0(n-2)=0\tag{2}
$$
is $R_0(n)=a_{-1}(-1)^n+a_33^n$, any solution of $(1)$ is unique modulo solutions of $(2)$.
One way to get a particular solution of $(2)$, create the matrix $M\in\mathbb{R}^{(n{+}1)\times(n{+}1)}$ satisfying
$$
M\begin{bmatrix}1\\x\\x^2\\\vdots\\x^n\end{bmatrix}=\begin{bmatrix}(I+S)(I-3S)1\\(I+S)(I-3S)x\\(I+S)(I-3S)x^2\\\vdots\\(I+S)(I-3S)x^n\end{bmatrix}\tag{3}
$$
Then, a particular solution of $(1)$ would be
$$
\begin{bmatrix}c_0&c_1&c_2&\cdots&c_n\end{bmatrix}M^{-1}\begin{bmatrix}1\\x\\x^2\\\vdots\\x^n\end{bmatrix}\tag{4}
$$
$M$ is lower triangular with $-4$s on the diagonal, therefore, invertible.
Another solution:
We can also derive a particular polynomial solution by noticing that Taylor's Theorem implies that, for polynomials at least,
$$
Sf(x)=e^{-D}f(x)\tag{5}
$$
where $D=\frac{\mathrm{d}}{\mathrm{d}x}$. $(5)$ is one of the motivations behind the Euler-Maclaurin Sum Formula.
Using $(5)$, $(1)$ becomes
$$
f(x)=(I-2e^{-D}-3e^{-2D})R(x)\tag{6}
$$
Noting that
$$
(1-2e^{-x}-3e^{-2x})^{-1}=-\tfrac{1}{4}-\tfrac{1}{2}x-\tfrac{9}{16}x^2-\tfrac{25}{48}x^3-\tfrac{15}{32}x^4-\dots\tag{7}
$$
we can get a partcular solution of $(1)$ with
$$
\begin{align}
R(x)
&=(1-2e^{-D}-3e^{-2D})^{-1}f(x)\\
&=(-\tfrac{1}{4}I-\tfrac{1}{2}D-\tfrac{9}{16}D^2-\tfrac{25}{48}D^3-\tfrac{15}{32}D^4-\dots)f(x)\tag{8}
\end{align}
$$