Partial fractions: different results from two methods

[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<br /><br /> -1=A+B<br /><br /> -3=A-B-C
\Rightarrow A=-\frac{5}{2}, B=\frac{3}{2}, C=-1<br />

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

 

[eumyang]

System of equations gives

-2=A+B+C<br /><br /> -1=A+B<br />-3=A-B-C
\Rightarrow A=-\frac{5}{2}, B=\frac{3}{2}, C=-1<br />

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

 

[bhoom]

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

Ahh yes. Thanks!