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Math Help - How To factor fractional exponents?????????

  1. #1
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    How To factor fractional exponents?????????

    Hey, I was having problems factoring/simplifying a problem I was given involving fractional and negative exponents

    This is the problem but I think I can figure out how to do it if you show me the general idea of factoring fractional exponents:

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


    Thanks in Advance
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  2. #2
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    Quote Originally Posted by Mudd_101 View Post
    Hey, I was having problems factoring/simplifying a problem I was given involving fractional and negative exponents

    This is the problem but I think I can figure out how to do it if you show me the general idea of factoring fractional exponents:

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


    Thanks in Advance
    1. The common denominator of the 2 summands is \sqrt{2x^2-x+1}

    2. From the sum in the numerator you can factor out (x^3+1)^{\frac13}

    3. Nevertheless the second factor looks like a nightmare
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  3. #3
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    Hey, uhh.... my math is really not very good so excuse any silly questions I may ask but uhh........ you said that both denominators are (sorry don't know how to put other people's quotes in)

    ((2x^2 - x +1)^1/2) but one of the ((2x^2 - x +1)^1/2) has a positive exponent, the other negative. I cannot combine two fractions, one with ((2x^2 - x +1)^(-1/2) and the other ((2x^2 - x +1)^(1/2) can I?


    tthanks
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  4. #4
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    Hello, Mudd_101!

    It looks like you were differentiating a product.
    If that's true, you've already made some errors . . .


     -\tfrac{1}{2}(2x^2-x+1)^{-\frac{1}{2}}(4x-1)(x^3+1)^{\frac{1}{3}} + (2x^2 -x + 1)^{\frac{1}{2}}\left(\tfrac{1}{3}\right)(x^3+1)(3  x^2)

    I assume the function was: . f(x) \:=\:\left(2x^2 - x + 1\right)^{\frac{1}{2}}\left(x^3+1\right)^{\frac{1}  {3}}

    Then: . f'(x) \;=\;\tfrac{1}{2}\left(2x^2-x+1\right)^{-\frac{1}{2}}(4x-1)\cdot\left(x^3+1\right)^{\frac{1}{3}} \;+\; (2x^2-x+1)^{\frac{1}{2}}\cdot\tfrac{1}{3}\left(x^3+1\rig  ht)^{-\frac{2}{3}}(3x^2)

    . . . . . f'(x)\;= \;\frac{(4x-1)(x^2+1)^{\frac{1}{3}}}{2(2x^2-x+1)^{\frac{1}{2}}} + \frac{x^2(2x^2-x+1)^{\frac{1}{2}}}{(x^3+1)^{\frac{2}{3}}}


    Get a common denominator:

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

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

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

    . . = \;\frac{8x^4 - 3x^3 + 2x^2 + 4x - 1}{2(x^3+1)^{\frac{2}{3}}(2x^2-x+1)^{\frac{1}{2}}}

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  5. #5
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    Ohhh!! Thanks Soroban I get it now.

    I will click that thanks thing for you.


    help was very appreciated bye.
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