Partial fractions: different results from two methods

1. Sep 25, 2011

bhoom

$$\frac{-2x^2-x-3}{x^3+2x^2-x-2}=\frac{-2x^2-x-3}{(x-1)(x+1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x+2}$$

Multiply by common denominator gives:

$$-2x^2-x-3=A(x+1)(x+2)+B(x-1)(x+2)+C(x-1)(x+1)\Leftrightarrow$$
$$-2x^2-x-3=Ax^2+A3x+A2+Bx^2+Bx-B2+Cx^2-C1$$

System of equations gives

$$-2=A+B+C$$$$-1=A+B$$$$-3=A-B-C$$
$$\Rightarrow A=-\frac{5}{2}, B=\frac{3}{2}, C=-1$$

However, "hand over" method gives:
$$[x=1]\Rightarrow A=\frac{ -2(1)^2-1-3}{(1+1)(1+2)}=\frac{-6}{6}=-1$$$$[x=-1]\Rightarrow B=\frac{ -2(-1)^2+1-3}{(-1-1)(-1+2)}=\frac{-4}{-2}=2$$$$[x=-2]\Rightarrow C=\frac{ -2(-2)^2+2-3}{(-2-1)(-2+1)}=\frac{-9}{3}=-3$$

Are both solutions correct, if so is that just a coincidence?
Have I made any errors, if so where?

2. Sep 25, 2011

eumyang

Looks like here's the problem. You forgot to include some integer coefficients. The system should be this:
$$A + B + C = -2$$$$3A + B = -1$$$$2A - 2B - C = -3$$

3. Sep 25, 2011

bhoom

Ahh yes. Thanks!