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Homework Help: 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


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    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]

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

    Do the factorization again.


    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.
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