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Math Help - binomial expansion

  1. #16
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    Re:

    RE:

    What? His text book says I was correct!
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  2. #17
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    Quote Originally Posted by qbkr21 View Post
    RE:

    What? His text book says I was correct!
    i'm afraid that method of expansion is incorrect. let's prove it.

    try you method of expansion to calculate:

    (1 + 2 + 3)(4 + 5)

    you will get 120

    the correct answer is 54


    or even look at the line where you have the expression for (e + f)^3 as e^3f + 2e^2f^2 + ef^3, you kow that's wrong
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  3. #18
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    Re:

    What?

    I got 54 too by the law of distribution
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  4. #19
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    Quote Originally Posted by qbkr21 View Post
    What?

    I got 54 too by the law of distribution
    yes, but look at how you expanded when you had:

    (e^2 + 2ef + f^2)(e + f)

    you did not use the law of distribution
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  5. #20
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    Here is doing it by hand:
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  6. #21
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    Re:

    RE:

    I just distributed one outside term to one inside term respected to order. I have done it for years this way on tests and have never had a problem. Your method would be the only way I could quickly go about solving very large binomials...
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  7. #22
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    Quote Originally Posted by qbkr21 View Post
    RE:

    I just distributed one outside term to one inside term respected to order. I have done it for years this way on tests and have never had a problem. Your method would be the only way I could quickly go about solving very large binomials...
    yeah, but after you expanded with respect to the first term, you then expanded with respect to the second term using the new terms you expanded using the first term, as opposed to the original terms. ok, that didn't make any sense. but look at your work again with e and f, and replace e and f with two numbers (preferably above 1), you'll realize that you don't get the right answer in the end
    Last edited by Jhevon; May 5th 2007 at 03:31 PM.
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  8. #23
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    Re:

    Ok
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  9. #24
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    Re:

    Jhevon I am so so sorry I was completely wrong! Just type in
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  10. #25
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    Quote Originally Posted by qbkr21 View Post
    Jhevon I am so so sorry I was completely wrong! Just type in
    wow, that expand feature is nice!

    you know, all this technology is going to spoil me one day. i don't even remember most of my time tables these days thanks to the calculator
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  11. #26
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    Re:

    I know. I for one started using a graphing calculator in the 6th grade. This has led to some serious problems...
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  12. #27
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    Quote Originally Posted by qbkr21 View Post
    I know. I for one started using a graphing calculator in the 6th grade. This has led to some serious problems...
    well, i never went that far...i don't even know how to use a graphing calculator. drawing graphs by hand has always been kind of enjoyable for me (i must sound like a real geek saying that, but i'm not) so i didn't let technology do it for me. but routine addition and subtraction and all that other stuff is just monotonous and painful, so i just whip out my calculator. i want to be able to use my head again and do this stuff, but i'm too far gone it seems, the habit sticks with me
    Last edited by Jhevon; May 5th 2007 at 12:39 PM.
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  13. #28
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    Hello, bobchiba!

    Does anyone know of an easy to remember way of expanding binomials?

    An example would be: .(x + y)^5
    First of all, we know what the terms look like:

    . . . x^5 . . x^4y . . xy . . xy . . xy^4 . . y^5 . . [1]

    It begins with x^5 and ends with y^5
    . . In between, the x-exponents decrease and the y-exponents increase.
    . . Note that, in each term, the exponents add up to 5.

    Now look at Pascal's Triangle and find the row that begins 1, 5.
    . . We find the numbers: .1, 5, 10, 10, 5, 1

    And those are the coefficients of the six terms in [1].

    Therefore: .(x + y)^5 .= .x^5 + 5x^4y + 10xy + 10xy + 5xy^4 + y^5

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Another example: .(2a + 3b)

    The terms are: . (2a) . . (2a)(3b) . . (2a)(3b) . . (3b)

    From Pascal's Triangle, we find: .1, 3, 3, 1

    So we have: .1(2a) + 3(2a)(3b) + 3(2a)(3b) + 1(3b)

    . . = .1(8a) + 3(4a)(3b) + 3(2a)(9b) + 1(27b)

    . . = .8a + 36ab + 54ab + 27b
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  14. #29
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    that last post really helped, but could you please just go over the last 2 lines again, thank you.
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