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Math Help - Partial Fraction Decomposition

  1. #1
    Newbie iEricKim's Avatar
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    Question Partial Fraction Decomposition

    Please explain how Partial Fraction Decomposition and how to solve for the constants for linear, repeated linear, and irreducible quadratic factors.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by iEricKim View Post
    Please explain how Partial Fraction Decomposition and how to solve for the constants for linear, repeated linear, and irreducible quadratic factors.
    this is a big question. too big for us to answer for you here. the first 5 or so links here should help. read up yourself and come back with specific questions. we offer math help here, not lessons
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  3. #3
    Newbie iEricKim's Avatar
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    Ok, then. Um, how do I turn a partial fraction decomposition into a system of equations to use as an augmented matrix to find the corresponding constants?

    Example problem:

    [ -(x^2) + 2x + 4 ] / [ x^3 - 4x^2 + 4x ]
    =
    A1/x + A2/(x-2) + A3/[ (x-2)^2 ]

    These are the decomposition factors, but what do I do now to find the constants?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by iEricKim View Post
    Ok, then. Um, how do I turn a partial fraction decomposition into a system of equations to use as an augmented matrix to find the corresponding constants?

    Example problem:

    [ -(x^2) + 2x + 4 ] / [ x^3 - 4x^2 + 4x ]
    =
    A1/x + A2/(x-2) + A3/[ (x-2)^2 ]

    These are the decomposition factors, but what do I do now to find the constants?
    \frac {-x^2 + 2x + 4}{x^3 - 4x^2 + 4x} = \frac Ax + \frac B{x - 2} + \frac C{(x - 2)^2}

    multiply both sides by the denominator of the left hand side:

    \Rightarrow -x^2 + 2x + 4 = A(x - 2)^2 + Bx(x - 2) + Cx

    let x = 0:~\Rightarrow 4 = 4A

    let x = 2:~ \Rightarrow 4 = 2C

    let x = 1:~\Rightarrow 5 = A - B + C

    there you go
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  5. #5
    Newbie iEricKim's Avatar
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    Where did you get the numbers 0, 2, and 1?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by iEricKim View Post
    Where did you get the numbers 0, 2, and 1?
    look at the expression we have. note that if x = 0 the terms multiplying B and C become zero, and hence i am just left with A to solve for. similarly, if x = 2, A and B go away, and i am left with C. x = 1 was a random number i used to get a third equation. i just picked something small enough to make computation easy. i would have already found two unknowns from the first two equations, so this last one will help me find the third
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  7. #7
    Newbie iEricKim's Avatar
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    Ah, ok! Thank you very much!
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