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Partial fractions: different results from two methods

  1. Sep 25, 2011 #1
    [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?
     
  2. jcsd
  3. Sep 25, 2011 #2

    eumyang

    User Avatar
    Homework Helper

    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]
     
  4. Sep 25, 2011 #3
    Ahh yes. Thanks!
     
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