Simplification of an Equation with Recurrence Relations I'm reading through examples on this site.
In example 2_2, given the recurrence relation $A_n - 2A_{n-1} = 2n^2$, the guess for the particular solution is $A_n= Bn^2 + Cn + D$. Substituting that into the recurrence relation gets you 
I just don't know how to continue this to get the answer above.

As Robert Israel pointed out, the following is obtained. 
Let $f_{n} = a n^{2} + b n + c$ in the difference equation $f_{n} = 2 f_{n-1} + 2 n^{2}$. 
\begin{align}
a n^{2} + b n + c &= 2 a (n-1)^{2} + 2b (n-1) + 2c + 2 n^{2} \\
&= 2a n^{2} - 4b n + 2a + 2b n - 2b + 2 n^{2} \\
&= 2(a+1) n^{2} + (b-2a) n + (a-b+c).
\end{align}
This leads to
\begin{align}
(a + 2) n^{2} + (b-4a) n + (2a - 2b + c) = 0.
\end{align}
Equating coefficients leads to $a = -2$, $b= 4a = -8$, and $c = 2b -2a = -12$ for which
\begin{align}
f_{n} = -2( n^{2} + 4n + 6) 
\end{align}
