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Math Help - Partial Fraction Decomposition Problem

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    Partial Fraction Decomposition Problem

    I'm trying to figure out this problem:


    At the moment, I've got it down to this:
    3x-4=A(x-1)+B(x-1)^2

    To find B, I let x=1, which also gets rid of A. Any ideas?
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  2. #2
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    Quote Originally Posted by BeSweeet View Post
    I'm trying to figure out this problem:


    At the moment, I've got it down to this:
    3x-4=A(x-1)+B(x-1)^2

    To find B, I let x=1, which also gets rid of A. Any ideas?
    Hi BeSweet,

    \frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}

    There must be a partial fraction present for each power of (x - c)^n less than or equal to n. We can find B, but not A using Heaviside's method. Only the constant corresponding to (x-c)^n can be found using this method.

    First, multiply the original proper fraction by (x-c)^2 and evaluate the result at x = c to get the value of the constant.

    \frac{3x-4}{(x-1)^2}(x-1)^2=3x-4 \Rightarrow B=3(1)-4=-1

    \boxed{B=-1}


    \frac{3x-4}{(x-1)^2}=\frac{A}{x-1}+\frac{-1}{(x-1)^2}

    To find the other constant, multiply both sides by the least common denominator to clear the fractions:

    3x-4=A(x-1)-1

    This leads to:

    3x-4=Ax + (-1-A)

    The polynomial on the left of the equal sign must have the same coefficient and constant as the polynomial on the right of the equal sign.

    \boxed{A=3}

    You could also say -4=(-1-A)

    You still get \boxed{A=3}

    The decomposition is:

    \boxed{\frac{3x-4}{(x-1)^2}=\frac{3}{x-1}+\frac{-1}{(x-1)^2}}
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