1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Factorizing & Manipulation

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    I need some help with this Algebra problem. In the following I don't know how they manipulated the RHS to get to the LHS:

    http://img402.imageshack.us/img402/7176/solns.gif [Broken]

    2. Relevant equations

    3. The attempt at a solution

    [tex]\frac{1}{4}(k+1)^2 .k^2 + (k+1)^3[/tex]

    [tex]\frac{1}{4}(k+1)^2 .k^2 + k^3 + 3k^2 +3k +1[/tex]

    I don't know how to manipulate [tex]k^2 + k^3 + 3k^2 +3k +1[/tex] into [tex](k^2+4k+4)[/tex]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 3, 2009 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Are you required to carry steps to change the right-side to be equal to the left-side? You should be allowed to carry steps on both sides so that you can show the right and left sides are equal to a third expression. The idea is, that if a=b, and if b=c, then a=c.
     
  4. Sep 3, 2009 #3
    another opinion of mine personally is to resolve 'complex' question initially is to model them by substitution.
    Like letting another alegrabic representation (eg) Let a = K+1 and then it will look simpler.

    Hope it helps.
     
  5. Sep 3, 2009 #4
    Well I'm not sure what they've done there. I mean how they simplified [tex][\frac{1}{2}k(k+1)]^2 + (k+1)^3[/tex] into [tex]\frac{1}{4} (k+1)^2 (k^2+4k+4)[/tex]

    Anyway, here's the rest (it's from a proof by induction problem):

    http://img200.imageshack.us/img200/6978/62815040.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
  6. Sep 3, 2009 #5
    [tex][\frac{1}{2}k(k+1)]^2 + (k+1)^3=(\frac{1}{2})^2k^2(k+1)^2+(k+1)(k+1)^2[/tex]

    Now just factor (k+1)2 and you are done. :approve:
     
  7. Sep 3, 2009 #6
    But that doesn't work! If I factor out the (k+1)2 I will have:

    [tex]\frac{1}{4}(k+1)^2 (k^2+k+1)[/tex]

    Which is not the same as:

    [tex]\frac{1}{4} (k+1)^2 (k^2+4k+4)[/tex]

    :uhh:
     
  8. Sep 3, 2009 #7
    You are wrong. :yuck:

    Do the factorization again.

    [tex]=(k+1)^2(\frac{1}{4}k^2+k+1)[/tex]

    Now factor 1/4 and see what will you come up with.
     
  9. Sep 3, 2009 #8
    Awww! Gee! I see what you mean now! Thanks a lot for the help.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Factorizing & Manipulation
  1. Factor This (Replies: 3)

  2. Factorize (Replies: 6)

  3. Math manipulation (Replies: 3)

  4. Algebraic Manipulation (Replies: 13)

Loading...