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  1. #1
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    partial fractions help,

    Express the following improper fractions as partial fractions.

    $\displaystyle \frac{x^{3} - x^{2} - x -3}{x(x-1)} $

    I know I am suppose to dive the numerator by the denominator, but I keep getting the wrong answer,
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  2. #2
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    Hello, Tweety!

    Express the following improper fraction as partial fractions:

    . . $\displaystyle \frac{x^3 - x^2 - x -3}{x(x-1)} $

    We have: .$\displaystyle \frac{x^3-x^2-x-3}{x^2-x} \;=\;x -\frac{x+3}{x(x-1)} $


    We will decompose the fraction:

    . . $\displaystyle \frac{x+3}{x(x-1)} \;=\;\frac{A}{x} + \frac{B}{x-1} \quad\Rightarrow\quad x+3 \;=\;A(x-1) + Bx $


    Let $\displaystyle x = 0\!:\quad0+3 \:=\:A(0 -1) + B(0) \quad\Rightarrow\quad A \:=\:-3$

    Let $\displaystyle x = 1\!:\quad 1+3 \:=\:A(1-1) + B(1) \quad\Rightarrow\quad B \:=\:4$


    The problem becomes: . $\displaystyle x - \left[-\frac{3}{x} + \frac{4}{x-1}\right] \;\;=\;\;x + \frac{3}{x} - \frac{4}{x-1}$

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  3. #3
    Junior Member slovakiamaths's Avatar
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    $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$
    now,
    $\displaystyle \frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$
    $\displaystyle x+3=A(x-1)+Bx$
    comparing coeff. of x and constant term we have, A=-3 & B=4
    so,
    $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]$
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  4. #4
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    Quote Originally Posted by slovakiamaths View Post
    $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$
    now,
    $\displaystyle \frac{x+3}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}$
    $\displaystyle x+3=A(x-1)+Bx$
    comparing coeff. of x and constant term we have, A=-3 & B=4
    so,
    $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}=x-[\frac{-3}{x}+\frac{4}{x-1}]$

    I am unsure as to how you got the first line? Could you please explain, how $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$

    thank you
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Tweety View Post
    I am unsure as to how you got the first line? Could you please explain, how $\displaystyle \frac{x^3-x^2-x-3}{x(x-1)}=x-\frac{x+3}{x(x-1)}$

    thank you
    factor the numerator $\displaystyle (x^3-x^2)-x-3 = x^2 (x-1) - x - 3 $

    So we get,

    $\displaystyle \frac{ x^2 (x-1) - x - 3 } { x(x-1) } = \frac{ x^2 (x-1) } { x(x-1) } - \frac { x + 3 } { x(x-1) } = x - \frac{x+3}{x(x-1)}$
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