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Math Help - Subtraction of fractions with polynomials

  1. #1
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    Subtraction of fractions with polynomials

    For a problem I'm working on, I get:
    \frac{1}{x}-\frac{2x}{x^{2}+1}

    My book has:
    \frac{1-x^{2}}{x(x^{2}+1)}

    Is this the same?
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  2. #2
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    e^(i*pi)'s Avatar
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    Yes, you can verify by cross multiplying

    \dfrac{1}{x} \cdot \dfrac{x^2+1}{x^2+1} = \dfrac{x^2+1}{x(x^2+1)}

    \dfrac{2x}{x^2+1} \cdot \dfrac{x}{x} = \dfrac{2x^2}{x(x^2+1)}


    Now the original expression can be rewritten as \dfrac{x^2+1}{x(x^2+1)} -  \dfrac{2x^2}{x(x^2+1)}

    Can you finish?
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Yes, you can verify by cross multiplying

    \dfrac{1}{x} \cdot \dfrac{x^2+1}{x^2+1} = \dfrac{x^2+1}{x(x^2+1)}

    \dfrac{2x}{x^2+1} \cdot \dfrac{x}{x} = \dfrac{2x^2}{x(x^2+1)}


    Now the original expression can be rewritten as \dfrac{x^2+1}{x(x^2+1)} -  \dfrac{2x^2}{x(x^2+1)}

    Can you finish?
    \dfrac{x^2+1}{x(x^2+1)} -  \dfrac{2x^2}{x(x^2+1)} =

    \dfrac{x^2+1-2x^2}{x(x^2+1)} =

    \dfrac{1-x^2}{x(x^2+1)}
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