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how do u use the binomial expansion formula? +ve rep for grab watch

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    say u had to expand (2x + 6)^8

    i know the first step is to write it out as 2(x+3)^8

    but then what do u do with all the formulas on this page: http://www.scribd.com/doc/15830460/E...t-New-Specimen

    i can do it manually via the triangle thingy, but i want to learn this method anyway.

    thanks in advance
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    (Original post by dumb maths student)
    say u had to expand (2x + 6)^8

    i know the first step is to write it out as 2(x+3)^8

    but then what do u do with all the formulas on this page: http://www.scribd.com/doc/15830460/E...t-New-Specimen

    i can do it manually via the triangle thingy, but i want to learn this method anyway.

    thanks in advance

    http://upload.wikimedia.org/math/d/3...69e01da47c.png

    So in this case n = 8, y = x and X = 3 (Excuse the bad notation)
    It's really just plug and play from there on and then at the end multiply everything you get by 2^8, since you took that factor out.
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    ahh right!

    When you factorise it, it comes out as:
    2^8 \times (x+3)^8

    Just for a while, forget the 28 part and concentrate on the (x+3)8.
    To expand this, you first off need to find the coefficients.
    There are terms involving x0, x1 ... x7, x8.

    The coefficient of x0 is given by:
    \frac{8!}{0! \times 8!} \times 3^8

    The coefficient of x1 is given by:
    \frac{8!}{1! \times 7!} \times 3^7

    The coefficient of x2 is given by:
    \frac{8!}{2! \times 6!} \times 3^6

    Can you see where this is going?

    It means that the first three terms in the expansion on (x+3)8 are:

    6561 + 17496x + 20412x^2

    Now to find your answer, there is the 28 term in there as well
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    (Original post by dumb maths student)
    say u had to expand (2x + 6)^8

    i know the first step is to write it out as 2(x+3)^8

    but then what do u do with all the formulas on this page: http://www.scribd.com/doc/15830460/E...t-New-Specimen

    i can do it manually via the triangle thingy, but i want to learn this method anyway.

    thanks in advance
    The 2 should be to the power 8

    2^8(x+3)^8
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    (Original post by placenta medicae talpae)
    ahh right!

    When you factorise it, it comes out as:
    2^8 \times (x+3)^8

    Just for a while, forget the 28 part and concentrate on the (x+3)8.
    To expand this, you first off need to find the coefficients.
    There are terms involving x0, x1 ... x7, x8.

    The coefficient of x0 is given by:
    \frac{8!}{0! \times 8!} \times 3^8

    The coefficient of x1 is given by:
    \frac{8!}{1! \times 7!} \times 3^7

    The coefficient of x2 is given by:
    \frac{8!}{2! \times 6!} \times 3^6

    Can you see where this is going?

    It means that the first three terms in the expansion on (x+3)8 are:

    6561 + 17496x + 20412x^2

    Now to find your answer, there is the 28 term in there as well
    oh.. so when you use that formula thingy do you just add them up? then multiply each coefficient by 2^8 at the end?
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    i remember being shown a much easier method to remember by my brother, where you literally just plug in numbers into a formula and the maths would do itself.

    anybody know what im talking about?
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    http://www2.warwick.ac.uk/services/e...sion/a2a5n.pdf

    general binomial expansion is what you're looking for? it's in pdf above.
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    (Original post by dumb maths student)
    oh.. so when you use that formula thingy do you just add them up? then multiply each coefficient by 2^8 at the end?
    Yep

    But - the expansion is only valid in the range:
    -\frac{1}{2} \le x \le \frac{1}{2}
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    (Original post by Keoje)
    http://www2.warwick.ac.uk/services/e...sion/a2a5n.pdf

    general binomial expansion is what you're looking for? it's in pdf above.
    YUSSSSS!
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    (Original post by dumb maths student)
    YUSSSSS!
    If you're using this, just ensure you're using the one which involves taking out a further 3^8 because it has to be of the form (y+1)n.

    And, in this case, (x+3)8 needs to be written as:
    3^8(\frac{x}{3}+1)^8
 
 
 
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