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B Factoring algebraic expressions contaning fractions

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  1. May 29, 2016 #1
    http://imgur.com/RNsBBoO (image)

    Can someone elaborate as to how he factored this? The textbook provided nothing else about factoring algebraic expressions with fractions than this image.It avoided fractions in factoring like a plague I guess.

    I understand the 3rd step to put the (x + 1)^(3/4) on the bottom since the exponent was negative. However I have absolutely no idea how the former 2 steps were possible.
     
  2. jcsd
  3. May 29, 2016 #2

    Mark44

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    Do you agree that ##(x + 1)^{1/4}## in the first step is the same as (equal to) ##(x + 1)^{-3/4}(x + 1)## in the second step?
    The purpose of doing this was to get a common factor of ##(x + 1)^{-3/4}##, and then using the distributive law.
     
  4. May 29, 2016 #3
    I have no idea how (x + 1)^(1/4) = (x + 1)^(-3/4) * (x + 1)

    EDIT: Ok, I get how it is equilavent
     
  5. May 29, 2016 #4
    In step 1 & 2, did he group factor?

    (2x + 1)(x + 1)^(-3/4) = (2x+1) / (x + 1)^(3/4)

    EDIT: OK no, he didn't group factor, he basically just added x and (x+1).
    I understand now, thanks.
     
  6. May 29, 2016 #5

    SteamKing

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    It's also equivalent. :wink:
     
  7. May 29, 2016 #6

    SteamKing

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    Naw. With a negative exponent, you can just re-write it so that term appears in the denominator.

    You know: ## a ⋅ b^{-n} = \frac{a}{b^n}##
     
  8. May 29, 2016 #7
    Yea, that was what I was thinking. Simply just add similar to adding fractions, since they have the same denominator, just add both numerators.
     
  9. May 29, 2016 #8

    SteamKing

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    You got it.
     
  10. May 29, 2016 #9
    Thanks both of you! :smile:
     
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