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Is there an easy way to simplify a 3 term squared paranthesis?

  1. Aug 25, 2010 #1
    I know I probably learned this 10 years ago, but I really have no idea how to simplify something like this:

    (A*B + C*D + E*F)2

    Is there any kind of shortcut (even a website that does it automatically!) that can help me solve expressions like this?
     
  2. jcsd
  3. Aug 25, 2010 #2
    what are A, B, C, D, and E? Real or complex?

    I'm not entirely sure what you mean by "simplify" because it already seems to be in simplified form. Did you mean expand? If you meant expand..

    Then use the general [tex] {(a +b)}^2 = a^2 +2ab +b^2 ...
    let AB+CD =a, and EF = b
    [/tex]
     
  4. Aug 25, 2010 #3
  5. Aug 25, 2010 #4
    I mean... there isn't really any shortcut to this (If thats what you are looking for)

    You can do (A*B + C*D + E*F)^2 = (A*B + C*D + E*F)*(A*B + C*D + E*F) and just multiply it out term by term.. or you can let a = AB+CD, and b = EF and use the general form for [tex] (a+b)^2 [/tex]

    [tex] so.. (AB + CD)^2 +2EF(AB +CD) + E^{2}F^2 = A^2B^2 + C^2D^2 +E^2F^2 + 2(ABCD + EFAB + EFCD) [/tex]

    I hope that's what you were looking for...or i'm not getting your question
     
  6. Aug 25, 2010 #5
    Thanks. That's what I tried to do (after your first post) but I got nervous, it didn't look like it was going to end up in the form that I wanted. I'll just multiply it out and see what I get though. All this time searching for a better way, I probably could have done it already.

    Thanks again.
     
  7. Aug 25, 2010 #6

    CRGreathouse

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  8. Aug 25, 2010 #7
    There is a trinominal formula, which can be generalized: (a+b+c)^3 = (3!/3!)[a^3+b^3+c^3) + 3!/2!1! [a^2(b+c)+b^2[a+c]+c^2[a+b]+3!/1!1!1! (abc).

    This can be applied to (a+b+c)^2 since there are only two kinds of answers m^2 or mn. The coefficients are 2!/2! or 2!/1!1!, which is 2 or 1, of course. We just rely on symmetry. a^2+b^2+c^2+[2](ab+ac+bc)].


    So this is rather easily handled by symmetry. More difficult cases can be handled, such as (a+b+c+d)^4. The coefficients are going to be like 4!/4! = 1, 4!/3!1! = 4, 4!/2!2! = 6, 4!/2!1!1!=12, and 4!/1!1!1!1! =24, the last coefficient being used in 24abcd.
     
    Last edited: Aug 25, 2010
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