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Math Help - factor completely

  1. #1
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    factor completely

    factor completely-math_image.aspx.gif


    factor this please
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: factor completely

    I'm having a hardtime reading the image.

    Is it

     \frac{1}{2}x^{\frac{-1}{2}}(3x+1)^{1/2} - \frac{3}{4}x^{\frac{1}{2}} (3x+1)^{\frac{-1}{2}} ?
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  3. #3
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    Re: factor completely

    yes it is
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  4. #4
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    Re: factor completely

    The image is a bit hard to read, but I presume you have:

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

    First, look at the constant factors...they have \frac{1}{4} as factors. Then look at the variable factors and pull out the factors with the smallest exponents.

    What do you get?
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  5. #5
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    Re: factor completely

    i get 1/4x^-1/2(3x+1)^-1/2
    how do you guys write it perfectly??
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: factor completely

    Quote Originally Posted by mathisfun26 View Post
    i get 1/4x^-1/2(3x+1)^-1/2
    how do you guys write it perfectly??
    We use a typesetting language called latex. Below is the way to write a simple equation in latex. Using Latex, you can write math and it is beautiful and easy to read

    The below link shows you how to use Latex.
    Introducing LaTeX Math Typesetting
    Last edited by jakncoke; November 24th 2012 at 06:54 PM.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: factor completely

    Yes, that's what will be out front, and what will be the factor left within parentheses?

    We are using \LaTeX to write the math expressions. Do a search here and online to find out how to use it. It is fairly straightforward.
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  8. #8
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    Re: factor completely

     \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}
    thanks!
    Last edited by mathisfun26; November 24th 2012 at 07:07 PM.
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  9. #9
    Senior Member jakncoke's Avatar
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    Re: factor completely

    Quote Originally Posted by mathisfun26 View Post
    <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{1}{4}x^{-1/2}(3x@plus;1)^{-1/2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}" title="\frac{1}{4}x^{-1/2}(3x+1)^{-1/2}" /></a>
    When ever you want to type latex start with [tex ] latex stuff [/ tex]. For example [tex ] \frac{1}{2} [/ tex].
    Thanks from mathisfun26
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: factor completely

    A neat feature here is that if you hover your mouse cursor over an expression written in LaTeX code, you will see the code used to generate the expression.
    Thanks from mathisfun26
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  11. #11
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    Re: factor completely

    \frac{1}{2} (3x+1) this is what i got when i divided it by the first term i'm not sure how to do it after that
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  12. #12
    MHF Contributor MarkFL's Avatar
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    Re: factor completely

    You are close...the first term within the parentheses would be:

    2(3x+1)

    What do you get for the second term?
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  13. #13
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    Re: factor completely

    why would it be 2??
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  14. #14
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    Re: factor completely

    i got -3x for the second
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  15. #15
    MHF Contributor MarkFL's Avatar
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    Re: factor completely

    It would be 2 because:

    \frac{\frac{1}{2}}{\frac{1}{4}}=2

    You are correct with the second term being -3x. So, put it all together, and what do you have?
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