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C2 BINOMIAL EXPANSION HELP!! pleaseeeee

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    Can somebody show me how to solve such equations? i just cant get my head around this :-
    1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.

    2. given that
    (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.

    Thank you
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    (Original post by sharon800)
    Can somebody show me how to solve such equations? i just cant get my head around this :-
    1. The coefficinet of x^3 in the expansion of (3+bx)^5 is -720. Find the value of the constant b.

    2. given that
    (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4, find the value of constants A,B and C.

    Thank you
    I use

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    For the first one you:

    1. Expand the brackets until you get ......x^3
    2. Make this coefficient equal to -720
    3. Simple algebra from here


    For the second one you:

    1. expand both to give the answer
    2. Total up the coeffients
    3. From here, work out what A, B, C are
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    Here is an example of the binomial

    (2+ax)^4 = 2^4 (1+\frac{ax}{2})^4 = 2^4[1 + 4\frac{ax}{2} + \frac{4*3}{2!}(\frac{ax}{2})^2 + \frac{4*3*2}{3!}(\frac{ax}{2})^3 + (\frac{ax}{2})^4]

    In my example the coefficient of x^2 is

    (2^4)(\frac{12a^2}{4})=48a^2
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    (Original post by Frodo Baggins)
    For the first one you:

    1. Expand the brackets until you get ......x^3
    2. Make this coefficient equal to -720
    3. Simple algebra from here


    For the second one you:

    1. expand both to give the answer
    2. Total up the coeffients
    3. From here, work out what A, B, C are
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong
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    (Original post by sharon800)
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2
    This is x^2

    and even if it were the correct term you have not squared your b
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    (Original post by sharon800)
    Thank you
    for the first one
    in the expansion of x^3 i got 5C3 *(3)^3*bx^2 = which is i think 270bx^3 = -720...and my answer is wrong...i dont know what i have done wrong
    I got b as -2

    Basically, to get x^3

    its 5c3 * (3)^2 * b^3x^3

    so 90b^3=-720

    so b^3=-8

    so b=-2

    i am unclear as to why i have got a neg. Is it cos im clever?
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    (Original post by Frodo Baggins)
    I got b as -2

    Basically, to get x^3

    its 5c3 * (3)^2 * b^3x^3

    so 90b^3=-720

    so b^3=-8

    so b=-2

    i am unclear as to why i have got a neg. Is it cos im clever?
    Thank you! i did this last nigh, and got it
    how do you do the second one?
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    (Original post by sharon800)
    Thank you! i did this last nigh, and got it
    how do you do the second one?
     \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

    Expand both  \displaystyle (2+x)^5 and  \displaystyle (2-x)^5 by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
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    (Original post by raheem94)
     \displaystyle (2+x)^5 (2-x)^5 = A+Bx^2+Cx^4

    Expand both  \displaystyle (2+x)^5 and  \displaystyle (2-x)^5 by using binomial expansion, then multiply both and compare the coefficients to find the value of A, B and C.
    i got A= 0? DOESNT sound right:/
    and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer?
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    (Original post by sharon800)
    i got A= 0? DOESNT sound right:/
    and the coefficents of b and c huge numbers? can you please solve this, so i can compare the answer?
    I get A as 1024. Try again.
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    (Original post by raheem94)
    I get A as 1024. Try again.
    lol i got 3125...:/
    how do you do it?
    i did 5C0*(5)^5*(-X)^0 and got -3125
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    (Original post by sharon800)
    lol i got 3125...:/
    how do you do it?
    i did 5C0*(5)^5*(-X)^0 and got -3125
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
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    (Original post by raheem94)
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
    thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
    i get it! thanks again!
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    (Original post by sharon800)
    thank you! i know where i went wrong...instead of putting 2^5 i put s^5 lol
    i get it! thanks again!
    No problem, you are welcome.
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    (Original post by raheem94)
     \displaystyle (2+x)^5 = ^5C_0 2^5(x)^0 + ^5C_12^4(x)^1...= 2^5 + 5\times 2^4x...=32+80x+...


     \displaystyle (2-x)^5 = ^5C_0 2^5(-x)^0 + ^5C_12^4(-x)^1...= 2^5 + 5\times 2^4(-x)...=32-80x+...

     \displaystyle (2+x)^5(2-x)^5=(32+80x+...)(32-80x+...)=32\times 32 +...=1024+...

    So  \displaystyle A=1024
    You told me once never to post full solutions. :confused:
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    (Original post by Frodo Baggins)
    You told me once never to post full solutions. :confused:
    Yes, it is not allowed to post full solutions.

    Full solutions are considered a last resort.

    Actually i didn't post the full solution, i only calculated one value, of A, to show the OP how to calculate them. He also had to calculate B and C, so it isn't a full solution, just a part of the solution to let the OP get started.

    Why do you think it to be a full solution?
    Have you read the OP's question?

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