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Math Help - Factor & Simplify

  1. #1
    Junior Member
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    Factor & Simplify

    I'm confused on how you solve this problem D;
    I don't know how to get the answer into a fraction without negative exponents.

    Help?

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

    A step by step explanation would really help me to understand it better.
    Thank you!
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, meiyukichan!

    There are several approaches to this problem.
    . . Here is one of them . . .


    Simplify: . -4x (3x - 2)^{-\frac{4}{3}} + 4(3x - 2)^{-\frac{1}{3}}
    Get rid of the negative exponents now . . .

    We have: . \frac{-4x}{(3x-2)^{\frac{4}{3}}} + \frac{4}{(3x-2)^{\frac{1}{3}}}


    Get a common denominator:

    . . \frac{-4x}{(3x-2)^{\frac{4}{3}}} + \frac{4}{(3x-2)^{\frac{1}{3}}}\cdot{\color{blue}\frac{3x-2}{3x-2}} \;\;=\;\; \frac{-4}{(3x-2)^{\frac{3}{2}}} + \frac{4(3x-2)}{(3x-2)^{\frac{3}{2}}}


    . . = \;\frac{-4x+4(3x-2)}{(3x-2)^{\frac{3}{2}}} \;\;=\;\;\frac{-4x + 12x - 8}{(3x-2)^{\frac{3}{2}}} \;\;=\;\;\boxed{\frac{8x-8}{(3x-2)^{\frac{3}{2}}}}

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  3. #3
    Junior Member
    Joined
    Sep 2008
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    Quote Originally Posted by Soroban View Post

    Thank you, thank you, thank you!
    Very helpful and now I understand that with negative exponents, you're suppose to write the inverse of it to get rid of the negative... right? heh.
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