Partial Fraction Expansion of $12\frac{x^3+4}{(x^2-1)(x^2+3x+2)}$

Question

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

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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!

Answer 1

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. $C=35$.