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Math Help - [SOLVED] help partial fractions

  1. #1
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    [SOLVED] help partial fractions

    hey i need help with this, X^2 + 9x + 8/(X^2 + x - 6)

    When I do the polynomial division I get 1 + 8x - 14. But then the answer for the partial fraction turns out all wrong when i check it against the answer giving in the textbook. The answer given in the book is : 1 + 2/(x - 3) + 6/(X + 1). How did they derive this answer? wtf am I doing wrong, its driving me nuts. Thank you.
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  2. #2
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    Quote Originally Posted by walk_in_fark View Post
    hey i need help with this, X^2 + 9x + 8/(X^2 + x - 6)

    When I do the polynomial division I get 1 + 8x - 14. But then the answer for the partial fraction turns out all wrong when i check it against the answer giving in the textbook. The answer given in the book is : 1 + 2/(x - 3)** + 6/(X + 1)**. How did they derive this answer? What am I doing wrong, its driving me nuts. Thank you.
    As you said

    1 + \frac{8x - 14}{x^2 + x - 6} = 1 + \frac{8x - 14}{(x+3)(x-2)}

    Now for the last piece look for a decomposition as follows

     \frac{8x - 14}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}

    multiply by

     (x+3)(x-2)

    giving

     8x - 14 = A(x-2) + B(x+3)

    expanding a group terms depending on whether there's a x or not

    A + B = 8
    2A - 3B = -14

    Solving for A and B gives A = 2, B = 6 so

     \frac{8x - 14}{(x+3)(x-2)} = \frac{2}{x+3} + \frac{6}{x-2}

    although it doesn't agree with what you have in your denominator (above **).
    Last edited by Jester; February 3rd 2009 at 05:03 AM. Reason: Fixed typo
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  3. #3
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    Thanks a lot, I now I know that is a type, again thank you.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by danny arrigo View Post
    As you said

    1 + \frac{8x - 14}{x^2 + x - 6} = 1 + \frac{8x - 14}{(x+3)(x-2)}

    Now for the last piece look for a decomposition as follows

     \frac{8x - 14}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}

    multiply by

     (x+3)(x-2)

    giving

     \color{red} 8x - 14 = A(x+2) + B(x-3)

    expanding a group terms depending on whether there's a x or not

    A + B = 8
    2A - 3B = -14

    Solving for A and B gives A = 2, B = 6 so

     \frac{8x - 14}{(x+3)(x-2)} = \frac{2}{x+3} + \frac{6}{x-2}

    although it doesn't agree with what you have in your denominator (above **).
    you made an error here. in the red line, you should have 8x - 14 = A(x - 2) + B(x + 3). you mixed the signs up. as it stands now, your solutions don't work. a minor mistake. the original poster should take note and fix the error him/herself
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