Results 1 to 6 of 6

Math Help - Simplifying an expression

  1. #1
    Newbie
    Joined
    Dec 2010
    Posts
    2

    Simplifying an expression

    Hello,
    I would like to know how to simplify this expression step-by-step:
    Wolfram Alpha

    If I try and do it by hand on paper, I get results like this:
    http://www.upload.ee/image/969390/clipboard_upped.png
    (I forgot to add bottom of fraction at the end, because top was already wrong, but the point of the problem remains...)
    Last edited by Wraith; December 6th 2010 at 03:37 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288
    Don't expand out until you have factorized as fully as possible.

    In this case, we have
    \frac{(m+n)(m^2-n^2)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}

    Note m^2-n^2=(m+n)(m-n)

    That leaves us with
     \frac{(m+n)(m-n)(m+n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}

     \frac{(m+n)^2-(m^2+mn+n^2)}{(m+n)}

    Answer should be visible now
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,867
    Thanks
    640
    \frac{m^2-n^2}{m-n} - \frac{m^3-n^3}{m^2-n^2}<br />

    common denominator ...

    \frac{(m+n)(m^2-n^2)}{(m+n)(m-n)} - \frac{m^3-n^3}{m^2-n^2}<br />

    \frac{(m+n)(m^2-n^2)-(m^3-n^3)}{(m+n)(m-n)}<br />

    factor the difference of squares and cubes ...

    \frac{(m+n)(m+n)(m-n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}<br />

    divide out the common factor (m-n) ...

    \frac{(m+n)(m+n)-(m^2+mn+n^2)}{m+n}

    expand (m+n)(m+n) ...

    \frac{(m^2+2mn+n^2)-(m^2+mn+n^2)}{m+n}

    combine like terms ...

    \frac{mn}{m+n}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Dec 2010
    Posts
    2
    Thank you, I understand that exercise now
    But, I came over this new one:



    What is the common denominator here? I could use {(x+2)(x-2) as a common denominator on first two blocks, but the last one has (x-2)^2} under it, how do I deal with that?

    This is where I am stuck..
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    692
    Hello, Wraith!

    \text{Simplify: }\;\dfrac{(m+n)(m^2-n^2) - (m-n)(m^2+mn + n^2)}{(m+n)(m-n)}

    We have: . \dfrac{(m^3 - mn^2 + m^2n - n^3) - (m^3 - n^3)}{(m+n)(m-n)}

    . . . . . . =\;\dfrac{m^3 - mn^2 + m^2n - n^3 - m^3 + n^3}{(m+n)(m-n)}

    . . . . . . =\;\dfrac{m^2n - mn^2}{(m+n)(m-n)}

    . . . . . . =\;\dfrac{mn(m-n)}{(m+n)(m-n)}

    . . . . . . =\;\dfrac{mn}{m+n}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,867
    Thanks
    640
    Quote Originally Posted by Wraith View Post
    Thank you, I understand that exercise now
    But, I came over this new one:



    What is the common denominator here? I could use {(x+2)(x-2) as a common denominator on first two blocks, but the last one has (x-2)^2} under it, how do I deal with that?

    This is where I am stuck..
    note that the common denominator is (x+2)(x-2)^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help simplifying expression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 1st 2011, 02:29 AM
  2. simplifying the expression
    Posted in the Algebra Forum
    Replies: 4
    Last Post: August 19th 2008, 10:20 AM
  3. I need help simplifying this expression...
    Posted in the Algebra Forum
    Replies: 4
    Last Post: March 17th 2008, 12:01 AM
  4. Replies: 1
    Last Post: February 24th 2008, 09:35 PM
  5. Simplifying an expression
    Posted in the Algebra Forum
    Replies: 6
    Last Post: July 7th 2007, 12:12 PM

Search Tags


/mathhelpforum @mathhelpforum