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Thread: Binomial Expansion

  1. #1
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    Binomial Expansion

    Expanding (2+h)^3

    I get 8h+8h^2+2h^3

    but...

    the answer is 8+12h+6h^2+h^3

    ...What am i doing wrong?
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  2. #2
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    Re: Binomial Expansion

    I can't say how you got it wrong, since all I see is your answer. The stated answer is correct.

    The binomial theorem for power 3 is this:

    (a + b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3

    To really highlight the pattern, see that that can be written as:

    (a + b)^3 = (1)a^3 b^0 + (3) a^2 b^1 + (3) a^1 b^2 + (1) a^0 b^3

    (That's because anything is 1 times itself, and because anything (non-zero) raised to the zero power is 1.)

    The pattern is: The powers of a start at 3 and drop by 1 with each term, ending at power = 0. The powers of b start at 0 and increase by 1 with each term, ending at power = 3.

    The coefficient pattern is 1, 3, 3, 1.

    Notice that the sum of the powers is 3 for each term. Notice that the formula is symmetric in order (because (a + b)^3 must equal (b + a)^3... and it does.)

    Examples:

    (a + 0)^3 = (a)^3 + 3 (a)^2 (0) + 3 (a) (0)^2 + (0)^3 = a^3 + 0 + 0 + 0 = a^3. (Check!)

    (a + (-a))^3 = (a)^3 + 3 (a)^2 (-a) + 3 (a) (-a)^2 + (-a)^3 = a^3 - 3a^3 + 3a^3 - a^3 = 0. (Check!)

    (x + 1)^3 = (x)^3 + 3 (x)^2 (1) + 3 (x) (1)^2 + (1)^3 = x^3 + 3x^2 + 3x + 1.

    (x - 1)^3 = (x)^3 + 3 (x)^2 (-1) + 3 (x) (-1)^2 + (-1)^3 = x^3 - 3x^2 + 3x - 1.

    (x + 2)^3 = (x)^3 + 3 (x)^2 (2) + 3 (x) (2)^2 + (2)^3 = x^3 + 6 x^2 + 12x + 8.

    (x - 2)^3 = (x)^3 + 3 (x)^2 (-2) + 3 (x) (-2)^2 + (-2)^3 = x^3 - 6 x^2 + 12x - 8.

    (5x - 4)^3 = (5x)^3 + 3 (5x)^2 (-4) + 3 (5x) (-4)^2 + (-4)^3 = 125x^3 - 300 x^2 + 240x - 64.

    (Check: 5^3 = 125, (3)(5)^2(-4) = -300, (3)(5)(-4)^2 = 240, (-4)^3 = -64

    (2 + 1)^3 = (2)^3 + 3 (2)^2 (1) + 3 (2) (1)^2 + (1)^3 = 8 + 3(4) + 6 + 1 = 8 + 12 + 7 = 20 + 7 = 27.

    (Note that 3^3 = 3(3)^2 = 3(9) = 27. Check!)

    (3 + 1)^3 = (3)^3 + 3 (3)^2 (1) + 3 (3) (1)^2 + (1)^3 = 27 + 3(9) + 9 + 1 = 27 + 27 + 10 = 54 + 10 = 64.

    (Note that 4^3 = 4(4)^2 = 4(16) = 64. Check!)

    (3 + 2)^3 = (3)^3 + 3 (3)^2 (2) + 3 (3) (2)^2 + (2)^3 = 27 + 3(9)(2) + 9(4) + 8 = 27 + 9(6) + 36 + 8 = 27 + 54 + 36 + 8 = 81 + 44 = 125.

    (Note that 5^3 = 5(5)^2 = 5(25) = 125. Check!)
    Last edited by johnsomeone; Oct 3rd 2015 at 05:00 PM.
    Thanks from macewan3135 and sakonpure6
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  3. #3
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    Re: Binomial Expansion

    $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

    try again ...
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  4. #4
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    Re: Binomial Expansion

    (h)^3 + 3(h)^2(2) + 3 (h)(2)^2 + (2)^3 = H^3 + 6h^2 +12h + 8

    ...

    Thank you!
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  5. #5
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    Re: Binomial Expansion

    Quote Originally Posted by macewan3135 View Post
    Expanding (2+h)^3

    I get 8h+8h^2+2h^3

    but...

    the answer is 8+12h+6h^2+h^3

    ...What am i doing wrong?
    As was pointed out, we can only point out mistakes if you show your work.

    Two ways to proceed

    First method: brute force.

    $(2 + h)^3 = (2 + h)^2(2 + h) = 2(2 + h)^2 + h(2 + h)^2 =$

    $2(4 + 4h + h^2) + h(4 + 4h + h^2) = 8 + 8h + 2h^2 + 4h + 4h^2 + h^3 = 8 + 12h + 6h^2 + h^3.$

    Brute force is not so hard with a power of 3, but it gets pretty messy with a power of 8 or 10.

    Second method: memorize the binomial expansion.

    $\displaystyle (a + b)^n = \sum_{i=0}^n\dbinom{n}{i}a^{(n-i)}b^i = \sum_{i=0}^n\dfrac{n!}{i! * (n - i)!}a^{(n-i)}b^i.$

    In which case you get the answer almost automatically simply by plugging in:

    $(2 + h)^3 = \dfrac{3!}{0! * 3!}2^{(3-0)}h^0 + \dfrac{3!}{1! * 2!}2^{(3-1)}h^1 + \dfrac{3!}{2! * 1!}2^{(3-2)}h^2 + \dfrac{3!}{3! * 0!}2^{(3-3)}h^3 \implies$

    $(2 + h)^3 = 2^3 + 3 * 2^2h + 3 * 2h^2 + h^3 = 8 + 12h + 6h^2 + h^3.$
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