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Math Help - Binomial theorem check

  1. #1
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    Binomial theorem check

    hey guys, just need to check if my working out and solution is correct.

    Find the coefficient of x^5 in (x^4 + 4)(3x-4)^9

    My solution

    So I started off by separating both brackets into 2 separate equations.

    1. x^4(3x-4)^9
    2. 4(3x-4)^9

    Now I find the coefficient of x in equation 1 and I find the coefficient of x^5 in equation 2.

    So: \binom{9}{9-i} (3x)^{9-i} (-4)^i

    now:

    1. \binom{9}{8} (3)^8 (-4)^1 = 9 \times 6561 \times (-4)

    2. 4 \binom{9}{4} (3)^4 (-4)^5 = 4 \times 126 \times 81 \times (-1024)

    So now adding both up:

    Solution: 9 \times 6561 \times (-4) + 4 \times 126 \times 81 \times (-1024)
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  2. #2
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    Hello jvignacio
    Quote Originally Posted by jvignacio View Post
    hey guys, just need to check if my working out and solution is correct.

    Find the coefficient of x^5 in (x^4 + 4)(3x-4)^9

    My solution

    So I started off by separating both brackets into 2 separate equations.

    1. x^4(3x-4)^9
    2. 4(3x-4)^9

    Now I find the coefficient of x in equation 1 and I find the coefficient of x^5 in equation 2.

    So: \binom{9}{9-i} (3x)^{9-i} (-4)^i

    now:

    1. \binom{9}{8} (3)^8 (-4)^1 = 9 \times 6561 \times (-4)

    2. 4 \binom{9}{4} (3)^4 (-4)^5 = 4 \times 126 \times 81 \times (-1024)

    So now adding both up:

    Solution: 9 \times 6561 \times (-4) + 4 \times 126 \times 81 \times (-1024)
    Your method is fine, but you have found the coefficient of x^8 in (1) and x^4 in (2).

    You need to write the terms the other way around:
    \big((-4)+3x\big)^9
    The term in x^i is now:
    \binom{9}{9-i}(-4)^{9-i}x^i
    Now you can put i = 1 and 5 respectively.

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello jvignacioYour method is fine, but you have found the coefficient of x^8 in (1) and x^4 in (2).

    You need to write the terms the other way around:
    \big((-4)+3x\big)^9
    The term in x^i is now:
    \binom{9}{9-i}(-4)^{9-i}x^i
    Now you can put i = 1 and 5 respectively.

    Grandad
    Ahh daym! no wonder I keep getting them wrong.. So its the other way around thank you Grandad :P
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello jvignacioYour method is fine, but you have found the coefficient of x^8 in (1) and x^4 in (2).

    You need to write the terms the other way around:
    \big((-4)+3x\big)^9
    The term in x^i is now:
    \binom{9}{9-i}(-4)^{9-i}x^i
    Now you can put i = 1 and 5 respectively.

    Grandad
    Hey Grandad, how do we know how what coefficient to get from 1 and 2.. In this example.. x^5(x^3+5)(5x+3)^8 , when you split the two brackets, you have to find the coeff of x^2 in the first bracket then find the coeff of x^5 in the second bracket.. but in the previous example it was x from 1 and x^5 from 2.

    thanks

    EDIT: NEVERMIND, JUST GOT IT
    Last edited by jvignacio; May 19th 2010 at 03:21 AM.
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