Page 2 of 2 FirstFirst 12
Results 16 to 22 of 22
Like Tree2Thanks

Math Help - factor completely

  1. #16
    Junior Member
    Joined
    Nov 2012
    From
    fefw
    Posts
    31

    Re: factor completely

    \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[3x+2] if the question said no negative exponents in the final answer how would it be?
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Junior Member
    Joined
    Nov 2012
    From
    fefw
    Posts
    31

    Re: factor completely

    \frac{3x+2}{4^{x^1/2} (3x+1)^{1/2}}  would it be this?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: factor completely

    You would have:

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

    Now simplify...what do you get?
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Junior Member
    Joined
    Nov 2012
    From
    fefw
    Posts
    31

    Re: factor completely

    \frac{1}{4}x^{-1/2}(3x+1)^{-1/2}[1] ??
    Follow Math Help Forum on Facebook and Google+

  5. #20
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: factor completely

    Yes, and you may omit the factor of 1. You may choose to write this as:

    \frac{1}{4\sqrt{x(3x+1)}}
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Junior Member
    Joined
    Nov 2012
    From
    fefw
    Posts
    31

    Re: factor completely

    Quote Originally Posted by MarkFL2 View Post
    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.
    Quote Originally Posted by jakncoke View Post
    When ever you want to type latex start with [tex ] latex stuff [/ tex]. For example [tex ] \frac{1}{2} [/ tex].
    [QUOTE=MarkFL2;754391]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.

    quick question how did u get 1/4 from the factors of 1/2 and 3/4
    Follow Math Help Forum on Facebook and Google+

  7. #22
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: factor completely

    \frac{1}{2}=\frac{1}{4}\cdot2 and \frac{3}{4}=\frac{1}{4}\cdot3
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Factor completely
    Posted in the Math Topics Forum
    Replies: 6
    Last Post: October 18th 2012, 09:34 AM
  2. factor each of the following completely
    Posted in the Algebra Forum
    Replies: 5
    Last Post: March 24th 2011, 01:56 PM
  3. Factor Completely
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2011, 09:20 PM
  4. Factor out COMPLETELY
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 30th 2010, 09:47 PM
  5. factor this completely
    Posted in the Algebra Forum
    Replies: 7
    Last Post: January 13th 2009, 11:49 PM

Search Tags


/mathhelpforum @mathhelpforum