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Math Help - Partial Fraction Decomposition

  1. #1
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    Partial Fraction Decomposition

    I thought i was doing fine until i ran into this problem:

    x^2+4x-8/x^3-x^2-4x+4

    soooo.... i factored out the bottom and got

    x^2+4x-8/(x+2)(x-2)(x-1)

    I have never encountered a problem like this one...

    Can any one help me out?

    Thank you, SO much.
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  2. #2
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    Find A,B,C such that

    \frac{x^2+4x-8}{(x+2)(x-2)(x-1)} = \frac{A}{x+2}+\frac{B}{x-2}+\frac{C}{x-1}
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  3. #3
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    Hello, Danktoker!

    I thought i was doing fine until i ran into this problem: . \frac{x^2+4x-8}{x^3-x^2-4x+4}

    So i factored and got: . \frac{x^2+4x-8}{(x-1)(x-2)(x+2)}

    I have never encountered a problem like this one. .
    Really? What's different about it?
    Standard set-up: . \frac{x^2+4x-8}{(x-1)(x-2)(x+2)} \;=\;\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x+2}

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  4. #4
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    Ive never had done one or seen an example of how to do one with

    X^3

    so i rly did not know how to set it up.

    Well imma give this a shot Ill report in a minute

    Thank you
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  5. #5
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    ok see this is why im stumped...

    ive never done one with that set up.


    Ok so far i got


    x^2+ 4x- 8=(A+ B+ C)x^2+ (-2B^2- 1B^2- 2C^2- 1C^2)x+ (-4A- 2B+ 2C)

    not to sure what to do from here :S
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Danktoker View Post
    ok see this is why im stumped...

    ive never done one with that set up.


    Ok so far i got


    x^2+ 4x- 8=(A+ B+ C)x^2+ (-2B^2- 1B^2- 2C^2- 1C^2)x+ (-4A- 2B+ 2C)

    not to sure what to do from here :S
    NO!!

    Look at what pickslides and soroban have suggested.

    You should definitely have these kind of examples in you book/notes. Otherwise Google is there.

    You can decompose the given term as:

    \frac{x^2+4x-8}{(x+2)(x-2)(x-1)} = \frac{A}{x+2}+\frac{B}{x-2}+\frac{C}{x-1}

    So,

    {x^2+4x-8} = A(x-2)(x-1) + B(x+2)(x-1) + C(x+2)(x-2).....(I)

    you have to use equation (I) to find the values of A,B,and C

    first,

    let x=1, then equation (I) becomes:

    1+4-8 = A(-1)(0) + B(3)(0)+ C(3)(-1)

    this gives: -3 = 0+0-3C \implies C=1

    Now set x = 2 and C=1 in equation (I) , and you'll find the value of B.

    Lastly, set x = -2, and you'll find the value of A.

    Thus you have the values of A,B and C and your partial fraction is decomposed.
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  7. #7
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    Thank you all very much,

    Sorry for appearing as a moron.

    But i had explained my situation before
    Our this a math 40 course, and our professor gives us take home "exams" with material from math 101-140 (sometime a lil higher) and this is what he grades us on. He actually straight up tells us that none of this stuff is in our books, but that he wants to weed out the weak ones....

    srsly, i think this guy is ****ing nuts, many complains have been filled on him, but no one seems to care......

    once again thank you all very much for helping me (and a few other of my classmates out with this stuff)
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  8. #8
    Math Engineering Student
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    \frac{x^{2}+4x-8}{(x+2)(x-2)(x-1)}=\frac{(x-2)(x+2)+4(x-1)}{(x+2)(x-2)(x-1)}=\frac1{x-1}+\frac{4}{(x-2)(x+2)}, and \frac{4}{(x-2)(x+2)}=\frac{(x+2)-(x-2)}{(x+2)(x-2)}=\frac{1}{x-2}-\frac{1}{x+2}.
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