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Algabra help

  1. May 12, 2003 #1
    (2X+Y)^5
    can someone tell me how to expand this?
     
  2. jcsd
  3. May 12, 2003 #2

    Tom Mattson

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    Just do successive multiplications:

    (2x+y)(2x+y)(2x+y)(2x+y)(2x+y)

    Multiply the first two:

    (4x2+4xy+y2)(2x+y)(2x+y)(2x+y)

    Then the next two, etc. It's messy, but straightforward.
     
  4. May 12, 2003 #3
    i know how to do that, but thanks for you replay. i am trying to find a faster way. possibly using factorials
     
  5. May 12, 2003 #4
    You use

    the Binomial Expansion, also known as Newtons Expansion.
     
  6. May 12, 2003 #5
    how do you do newtons expansion?
     
  7. May 13, 2003 #6
    Do you know Pascal Triangle? Or have you learnt combinations, Cnr, before?
     
  8. May 13, 2003 #7
    Newton's binomial (a.k.a. Newton's expansion) is this:

    (a+b)^n=(a^n)+(n*((a^(n-1))*b))+((n*(n-1)*(a^(n-2))*b^2)/(2!))+((n*(n-1)*(n-2)*(a^(n-3))*b^3)/(3!))+...+(b^n)

    n can be any rational number
     
    Last edited: May 13, 2003
  9. May 17, 2003 #8
    Or shorter

    (a+b)^n=SUM (from m=0 to n) C(m out of n)*a^m*b^(n-m)

    Damn can I turn on the HTML code?
     
    Last edited by a moderator: May 17, 2003
  10. May 22, 2003 #9
    Perhaps I can help you with that.

    This is how it goes from a simple binomial theorem:

    (a+b)0=1 since anything to the power of zero is 1

    (a+b)1=a+b

    (a+b)2=a2+2ab+b2

    (a+b)3=a3+3a2b+3ab2+b3

    As we go on and on we can clearly see that a pattern is emerging. Look at the next post just following this.
     
  11. May 22, 2003 #10
    First, notice that if we add the powers for a and b the result is always equal to the original power given to the term.

    For example in (a+b) 2= a 2+2ab+b 2 notice that power in each term is always equal to 2. The first a2, the second a 1 and b 1 and again 1+1 is equal 2. The same goes for a3+3a2b+3ab3+b3 and on and on.
    Go to the next coming post.
     
  12. May 22, 2003 #11
    Second, it is apparent that power decreases from a, and increases in b as we go forward.

    For example in (a+b)5=a 5 b 0+ a 4 b 1 + a 3 b 2 + a 2 b 3 + a 1 b 4 + a 0 b 5.
    As it must become obvious from the above, a starts with power 5 and goes to power 0 and b starts with power 0 and goes to power 5. Of course in the above we are missing the coefficient for each term. Now I show you how to find them.
     
  13. May 22, 2003 #12
    To make writing the coefficients clear I rewrite the above powering using only coefficients.

    i.e. how many of each kind of term:
    (a + b)
    1 1 0+
    0 1 1
    -------
    1 2 1 0
    (a + b)2
    1 2 1 0+
    0 1 2 1
    ----------
    (a + b) 3
    1 3 3 1 0 +
    0 1 3 3 1
    -------------
    (a + b) 4
    1 4 6 4 1 +

    0 1 4 6 4 1
    ---------------------------
    a + b) 5
    1 5 10 10 5 1

    This is what is known as Pascal's Triangle. The last thing that you have to do is substitute 2x for a and y for b in the above. Good luck
    :wink:
     
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