Partial Fraction Expansion of $12\frac{x^3+4}{(x^2-1)(x^2+3x+2)}$ Find the vector $(A,B,C,D)$ if $A$, $B$, $C$, and $D$ are the coefficients of the partial fractions expansion of
$$12\frac{x^3+4}{(x^2-1)(x^2+3x+2)} = \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}.$$

After plugging in a few values for $x,$ I got these equations:
$$6A+D=12$$
$$4A+B+2C+D=0$$
$$5A+D=B+C$$
$$2A+B=2C+2D+48$$
But now I'm stuck.  Thanks in advance for answering!

You should not begin with solving systems of linear equations. Rewrite the partial fractions decomposition so as to remove all denominators. Multiplying both sides with $(x-1)(x+2)(x+1)^2$, you get
$$12(x^3+4)=A(x+2)(x+1)^2+B(x-1)(x+1)^2 +C(x-1)(x+2)(x+1)+D(x-1)(x+2).$$ 
Now set 


*

*$x=1$: it  yields at once $12\cdot 5= 12A$, whence $A=5$;

*$x=-2\;$ yields $\;12^2=-3B$, whence $B=-48$;

*$x=-1\;$ yields $\;12\cdot 5=-2D$, whence $D=-30$.


There remains to calculte $C$. The simplest is to set $x=0$, and you get $48=2A-B-2C-2D$, i.e. $C=35$.
