Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$. 
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.

Put $x=1$,
$3=9A$ $\implies$ $A=\dfrac13$.
\begin{align*}
x^4-x^3+2x^2-x+2 &= \frac13(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\
\frac23x^4-x^3+\frac23x^2-x+\frac23&=  (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)\\
\frac23x^3-\frac13x^2+\frac13x-\frac23&=  (Bx+C)(x^2+2) + Dx+E\\
\end{align*}
By division, $\dfrac23x^3-\dfrac13x^2+\dfrac13x-\dfrac23=(x^2+2)\left(\dfrac23 x-\dfrac13\right)-x$.
