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Math Help - Complex Fractions & Negative Exponents

  1. #1
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    Complex Fractions & Negative Exponents

    So I've set a goal for myself to relearn all the math I once knew, one textbook at a time. Realizing I could do at age 10 what I can't do now is humbling. Anyway.

    I've got this problem and can't figure out how to get the correct answer:

    c-2 - 1
    -------
    c-1 + 1

    I've tried to work it as follows:


    1
    -- - 1
    c2

    ---------

    1
    --- + 1
    c

    Clear the fraction by multiplying by c2 across all numerators.

    This leaves


    1 - c2

    -------
    c + c2


    However, my book says the answer is:

    1 - c
    ------
    c

    I can't for the life of me get there.

    Help?

    [[EDIT:
    Oh good lord. After trying for over an hour I worked backwards from the given solution and got how to simplify to the correct answer.

    However, if this had been a proper exam I would've never thought to factor down to what is 'correct'. I only got there because I knew what was right.

    Any tips on knowing when factoring is possible? Or quicker ways to recognize those patterns? Help would be appreciated.]]
    Last edited by pigeonpierce; July 28th 2013 at 10:55 PM.
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  2. #2
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    Re: Complex Fractions & Negative Exponents

    Hey pigeonpierce.

    In terms of advice for factoring: try and reduce expressions down to the smallest factors possible.

    The simplest factor possible is a multiplication of linear factors like (1+c)(1-c). If you can't get linear factors, then just go as far as you can towards linear.

    The rest will be cancelling out factors leaving you with the minimal representation.
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  3. #3
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    Re: Complex Fractions & Negative Exponents

    Quote Originally Posted by pigeonpierce;793798I've got this problem and can't figure out how to get the correct answer:

    c[SUP
    -2 [/SUP]- 1
    -------
    c-1 + 1

    We can factor this simply by using the difference of two squares.

    \frac{c^{-2}-1}{c^{-1}+1}=\frac{(c^{-1}-1)(c^{-1}+1)}{c^{-1}+1}=c^{-1}-1
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  4. #4
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    Re: Complex Fractions & Negative Exponents

    Quote Originally Posted by pigeonpierce View Post
    So I've set a goal for myself to relearn all the math I once knew, one textbook at a time. Realizing I could do at age 10 what I can't do now is humbling. Anyway.

    I've got this problem and can't figure out how to get the correct answer:

    c-2 - 1
    -------
    c-1 + 1

    I've tried to work it as follows:


    1
    -- - 1
    c2

    ---------

    1
    --- + 1
    c

    Clear the fraction by multiplying by c2 across all numerators.

    This leaves


    1 - c2

    -------
    c + c2
    Plato has already indecated that you could use the "difference of squares" factoring directly from the original statement of the problem. You can also use it here:
    \frac{1- c^2}{c+ c^2}= frac{(1- c)(1+ c)}{c(1+ c)}= \frac{1- c}{c}
    since we can cancel the two "1+ c".


    However, my book says the answer is:

    1 - c
    ------
    c

    I can't for the life of me get there.

    Help?

    [[EDIT:
    Oh good lord. After trying for over an hour I worked backwards from the given solution and got how to simplify to the correct answer.

    However, if this had been a proper exam I would've never thought to factor down to what is 'correct'. I only got there because I knew what was right.

    Any tips on knowing when factoring is possible? Or quicker ways to recognize those patterns? Help would be appreciated.]][/QUOTE]
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