Partial fractions: different results from two methods

[bhoom]

[tex]\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}[/tex]


Multiply by common denominator gives:

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

System of equations gives

[tex]-2=A+B+C
[/tex][tex]
-1=A+B
[/tex][tex]
-3=A-B-C[/tex]
[tex]\Rightarrow A=-\frac{5}{2}, B=\frac{3}{2}, C=-1
[/tex]

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

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

 

[eumyang]

System of equations gives

[tex]-2=A+B+C
[/tex][tex]
-1=A+B
[/tex][tex]-3=A-B-C[/tex]
[tex]\Rightarrow A=-\frac{5}{2}, B=\frac{3}{2}, C=-1
[/tex]

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

 

[bhoom]

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

Ahh yes. Thanks!