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Expansion of (x^2 - 2)^5

  1. Aug 21, 2006 #1
    This is a question from a test that I have done. I have the answer but am unsure of the working.

    consider the expansion of (x^2 - 2)^5
    a) write the down the number of terms in this expansion
    b) the first 4 terms of the expansion in descending powers of x are

    x^10 - 10x^8 + 40x^6 + Ax^4 + ...

    Find the value of A

    For part a) I got the answer as 5 but apparantly it is 6 and I am not sure why.

    I am stuck how to do part b) but know the answer is -80
     
  2. jcsd
  3. Aug 21, 2006 #2
    Well you can either use the binomial theorem, or multiply it all out yourself and see why there are 6 terms and A is -80.
     
  4. Aug 28, 2006 #3
    Do you know pascals triangle? If you look down to the sixth row there are 6 numbers, hence your expansion has 6 terms. You can think of it like this: a quadratic has 3 terms, a cubic has 4 terms etc. As a rule if you have an expansion to the power n there will be n+1 terms (assuming none of the terms cancel out.)

    For part B you will have to use the binomial expansion. You will need to use pascals triangle or "choose" notation to calculate the multiplier and then use the fact that the sum of the powers of each term on the right needs to equal the power on the left (IE if the x^2 is cubed then the -2 needs to be squared.) With these two things you should calculate the right answer.


    :)
     
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