Results 1 to 4 of 4

Math Help - Simplify the below equation

  1. #1
    Newbie
    Joined
    Feb 2011
    Posts
    17

    Post Simplify the below equation

    (k-4)/(K^2+5K+6) * (k^2+8K+12)/(K^2-10K+24)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    You can factor all those expressions to make life simpler

    \dfrac{(k-4)}{(k+3)(k+2)} \cdot \dfrac{(k+6)(k+2)}{(k-6)(k-4)}

    As long as k \neq -2, -3, 4, 6 you can do some cancelling
    Last edited by e^(i*pi); February 19th 2011 at 09:34 AM. Reason: expressions, not equations
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Deleted by repetition.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    That is not an equation but a rational fraction. To simplify, decompose the second degree polynomials, for example k^2-10k+24=(k-4)(k-6) .


    Fernando Revilla


    Edited: Sorry, I didn`t see e^(i*pi)'s post.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How would you simplify this equation?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 12th 2010, 01:26 AM
  2. Simplify Equation
    Posted in the Algebra Forum
    Replies: 3
    Last Post: January 30th 2010, 07:09 PM
  3. simplify equation
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 13th 2009, 10:53 PM
  4. How would I simplify these equation?
    Posted in the Algebra Forum
    Replies: 8
    Last Post: December 13th 2009, 10:38 PM
  5. Simplify the equation
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 9th 2005, 12:13 PM

Search Tags


/mathhelpforum @mathhelpforum