Results 1 to 4 of 4

Math Help - Quick Question on Partial Fractions Problem

  1. #1
    Super Member
    Joined
    Dec 2008
    Posts
    509

    Quick Question on Partial Fractions Problem

    Hi
    Can someone tell me if my answer is correct or my book's answers

    1) Resolve into a partial fraction: \frac{x+1}{(x^2+4)(x^2-1)}

    i got \frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}

    however book's answers is  \frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}<br />
    P.S
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone tell me if my answer is correct or my book's answers

    1) Resolve into a partial fraction: \frac{x+1}{(x^2+4)(x^2-1)}

    i got \frac{\frac{1}{5}}{(x-1)}-\frac{\frac{1}{5}}{(x^2+4)}

    however book's answers is  \frac{\frac{1}{5}}{(x-1)}-\frac{(x+1)}{5(x^2+4)}<br />
    P.S
    The book's answer is correct.

    \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}

     = \frac{(Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4)}{(x^2 + 4)(x^2 - 1)}.


    So (Ax + B)(x^2 - 1) + (Cx + D)(x^2 + 4) = x + 1

    Ax^3 - Ax + Bx^2 - B + Cx^3 + 4Cx + Dx^2 + 4D = x + 1

    (A + C)x^3 + (B + D)x^2 + (4C - A)x + 4D - B = 0x^3 + 0x^2 + 1x + 1.


    So A + C = 0

    B + D = 0

    4C - A = 1

    4D - B = 1


    Solving these equations simultaneously gives

    A = -\frac{1}{5}, B = -\frac{1}{5}, C = \frac{1}{5}, D = \frac{1}{5}.


    So \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{-\frac{1}{5}x - \frac{1}{5}}{x^2 + 4} + \frac{\frac{1}{5}x + \frac{1}{5}}{x^2 - 1}

     = -\frac{x + 1}{5(x^2 + 4)} + \frac{x + 1}{5(x^2 - 1)}

     = \frac{x + 1}{5(x - 1)(x + 1)} - \frac{x + 1}{5(x^2 + 4)}

     = \frac{1}{5(x - 1)} - \frac{x + 1}{5(x^2 + 4)}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Dec 2008
    Posts
    509
    Quote Originally Posted by Prove It View Post
    The book's answer is correct.

    \frac{x + 1}{(x^2 + 4)(x^2 - 1)} = \frac{Ax + B}{(x^2 + 4)} + \frac{Cx + D}{(x^2 - 1)}
    yeh i made it  = \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 +  4)}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,656
    Thanks
    1480
    Quote Originally Posted by Paymemoney View Post
    yeh i made it  = \frac{A}{(x-1)} + \frac{B}{(x+1)}+\frac{C}{(x^2 +  4)}.
    When dealing with partial fractions, it's always easiest to work with two factors, and then break it down further later if possible.

    Also, your way wouldn't work as it should actually be

    \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{Cx + D}{x^2 + 4}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. partial fractions question
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: September 14th 2010, 11:47 AM
  2. Partial Fractions Question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: March 26th 2010, 04:24 AM
  3. Quick Question (fractions and exponents)
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 3rd 2009, 01:56 PM
  4. Quick question about fractions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 30th 2008, 10:39 AM
  5. Replies: 1
    Last Post: August 31st 2007, 12:47 PM

Search Tags


/mathhelpforum @mathhelpforum