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Math Help - Please help with Partial Fractions

  1. #1
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    Please help with Partial Fractions

    Hi, I'm working on the decomposition of this partial fraction and I solved for two of the constant. I cant seem to solve for the constant "A". Can someone please help me with this one?

    Thanks,

    Coop
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by coopsterdude View Post
    Hi, I'm working on the decomposition of this partial fraction and I solved for two of the constant. I cant seem to solve for the constant "A". Can someone please help me with this one?

    Thanks,

    Coop
    I'd do it this way (my preferred method):
    4x^2 + 2x - 1 = A(x^2 + x) + B(x + 1) + Cx^2

    4x^2 + 2x - 1 = (A + C)x^2 + (A + B)x + B

    So matching coefficients we get
    A + C = 4

    A + B = 2

    -1 = B
    which you can solve easily.

    -Dan
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  3. #3
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    Hello, Coop!

    I've never seen a layout like yours.
    . . [I wonder where your teacher (or book) learned that approach.]
    I've seen topsquark's method many times.

    I'll show you the method I was taught . . .


    \frac{4x^2+2x-1}{x^2(x+1)} \;=\;\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1}

    Multiply through by the LCD: . 4x^2 + 2x-1 \;=\;x(x+1)A + (x+1)B + x^2C


    Now we select some "good" values for x.

    Let x = 0:\;\;\text{-}1 \:=\:0\!\cdot\!A + 1\!\cdot\!B + 0\!\cdot\!\quad\Rightarrow\quad\boxed{B \,=\,\text{-}1}

    Let x = \text{-}1:\;\;1 \:=\:0\!\cdot\!A + 0\!\cdot\!B + 1\!\cdot\!C\quad\Rightarrow\quad\boxed{C \,=\,1}

    Let x = 1:\;\;5 \:=\:2A + 2B + C
    . . Since B = \text{-}1,\:C = 1, we have: . 5 \:=\:2A - 2 + 1\quad\Rightarrow\quad\boxed{A \,=\, 3}


    Therefore: . \frac{4x^2+2x-1}{x^2(x+1)} \;=\;\frac{3}{x} - \frac{1}{x^2} + \frac{1}{x+1}

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  4. #4
    Super Member wingless's Avatar
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    Well, I think he did the same thing. He just multiplied every term by the denominator and wrote it, instead of writing the result directly.
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