Results 1 to 3 of 3

Math Help - Probability Probelms & Counting Techniques

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    13

    Probability Probelms & Counting Techniques

    Expand using the binomial theorem. Simplify the result fully.

    (x^2 - 3y)^4
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    (x^2)^4 + 4\{(x^2)^3 \cdot (-3y)^1\} + 6\{(x^2)^2 \cdot (-3y)^2\} + 4\{(x^2)^1 \cdot (-3y)^3\} + (-3y)^4 \\

    \Rightarrow x^8 + 4(x^6 \cdot -3y) +6(x^4 \cdot 9y^2) + 4(x^2 \cdot -27y^3) + 81y^4

    ...

    Ouch, I'm pretty tired- thanks Soroban.
    Last edited by Nyoxis; June 10th 2008 at 05:00 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    693
    Hello, 16sammy

    Expand using the binomial theorem: . (x^2 - 3y)^4
    We have: . (x^2-3y)^4

    . . = \;\;(x^2)^4 \;+ \;4(x^2)^3(\text{-}3y)^1 \;+\; 6(x^2)^2(\text{-}3y)^2 \;+\; 4(x^2)^1(\text{-}3y)^3 \;+\; (\text{-}3y)^4

    . . = \;\;x^8 \;+\; 4(x^6)(\text{-}3y) \;+ \;6(x^4)(9y^2) \;+ \;4(x^2)(\text{-}27y^3) \;+ \;81y^4

    . . = \;\;x^8 \;- \;12x^6y \;+ \;54x^4y^2 \;- \;108x^2y^3 \;+ \;81y^4

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Counting Techniques.
    Posted in the Statistics Forum
    Replies: 3
    Last Post: October 24th 2009, 03:14 PM
  2. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 11
    Last Post: June 10th 2008, 08:03 PM
  3. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 10th 2008, 05:21 PM
  4. Probability Probelms & Counting Techniques
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 10th 2008, 04:52 PM
  5. probability problem counting techniques.
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: September 27th 2007, 07:50 AM

Search Tags


/mathhelpforum @mathhelpforum