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Math Help - Binomial expansion help- is the question wrong?

  1. #1
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    Binomial expansion help- is the question wrong?

    Find the coefficiant of x in the expansion of (1 - 2x)^4 (2 + x)^9
    Ive done this (I think) and got:

    512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9

    I cannot find a coeficciant of x because, after expanding the 2 brackets. Upon multiplying them both together, the 2 x's that are in each bracket make an x^2 therefore eradicating a coefficiant of x

    Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

    thanks!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by fishkeeper View Post
    Ive done this (I think) and got:

    512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9

    I cannot find a coeficciant of x because, after expanding the 2 brackets. Upon multiplying them both together, the 2 x's that are in each bracket make an x^2 therefore eradicating a coefficiant of x

    Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

    thanks!
    You should know that you don't have to do the full expansions for this type of question?

    (1-2x)^4(2+x)^9=[1+4(-2x) + ... ]2^9[1+9(x/2) +...]=2^9[1-8x+(9/2)x+...]

    where the  ... denote terms of higher order than 1

    So the coefficient you are looking for is: -2^9(7/2)=-7 \times 2^8

    CB
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  3. #3
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    Quote Originally Posted by fishkeeper View Post
    Ive done this (I think) and got:

    512 - 18432 x^2 + 110592 x^4 + 2016 x^5 - 171360 x^6 + 144 x^7 + 64521 x^8 + x^9

    I cannot find a coeficciant of x because, after expanding the 2 brackets. Upon multiplying them both together, the 2 x's that are in each bracket make an x^2 therefore eradicating a coefficiant of x

    Any help would be greatly appreciated as if this comes up in the exam, I would be prety stuck

    thanks!
    Hi fishkeeper,

    (1-2x)^4=1+\binom{4}{1}(-2x)+\binom{4}{2}(-2x)^2+....last\ 2\ terms

    (2+x)^9=2^9+\binom{9}{1}2^8x+\binom{9}{2}2^7x^2+..  .remaining\ terms

    The coefficient of x is found from the constant in one expanded factor multiplied by the coefficient of x in the other.

    This is (1)\binom{9}{1}2^8\ added\ to\ \binom{4}{1}(-2)2^9
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    You should know that you don't have to do the full expansions for this type of question?

    (1-2x)^4(2+x)^9=[1+4(-2x) + ... ]2^9[1+9(x/2) +...]=2^9[1-8x+(9/2)x+...]

    where the  ... denote terms of higher order than 1

    So the coefficient you are looking for is: -2^9(7/2)=-7 \times 2^8

    CB

    Sorry, but I really don't understand this at all.

    Why does it not have to be expanded? And, are you just putting the 2 brackets into the general formula for an arithmetic progression? As it looks vaguely like that?
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  5. #5
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    Hi fishkeeper,

    Higher powers of x will not be involved in computing the x^1 term, since multiplications of those will only contribute to higher powers of x.

    For instance

    (3+x+x^2)(2)=6+2x+2x^2

    The x term results from the multiplication of x by the constant.

    Or..

    (3+x+x^2)(2+x+x^2)=6+3x+3x^2+2x+x^2+x^3+2x^2+x^3+x  ^4

    =6+(3x+2x)+(3x^2+x^2)+(x^3+x^3)+x^4

    The resultant x term is the sum of the constant from one factor multiplied by the x term in the other factor from both sets of factors.

    When CaptainBlack wrote out the pair of square brackets multiplied together,
    he only wrote out the first few terms of the binomial expansions, as higher powers of x will not contribute to the x^1 term.
    That's all the question wanted you to discover.
    The question wanted you to see that only a pair of terms from both expansions contribute to the resultant x term.

    CaptainBlack placed the first few terms of the binomial expansions of both (1-2x)^4 and (2+x)^9 into the square brackets, as it is unnecessary to write the rest of them to obtain the coefficient of the x^1 term only.
    if you needed the entire expansion, then all the terms of both would be required.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Archie Meade View Post
    Hi fishkeeper,

    Higher powers of x will not be involved in computing the x^1 term, since multiplications of those will only contribute to higher powers of x.

    For instance

    (3+x+x^2)(2)=6+2x+2x^2

    The x term results from the multiplication of x by the constant.

    Or..

    (3+x+x^2)(2+x+x^2)=6+3x+3x^2+2x+x^2+x^3+2x^2+x^3+x  ^4

    =6+(3x+2x)+(3x^2+x^2)+(x^3+x^3)+x^4

    The resultant x term is the sum of the constant from one factor multiplied by the x term in the other factor from both sets of factors.

    When CaptainBlack wrote out the pair of square brackets multiplied together,
    he only wrote out the first few terms of the binomial expansions, as higher powers of x will not contribute to the x^1 term.
    That's all the question wanted you to discover.
    The question wanted you to see that only a pair of terms from both expansions contribute to the resultant x term.

    CaptainBlack placed the first few terms of the binomial expansions of both (1-2x)^4 and (2+x)^9 into the square brackets, as it is unnecessary to write the rest of them to obtain the coefficient of the x^1 term only.
    if you needed the entire expansion, then all the terms of both would be required.
    Another point is that a full expansion could be impractical but the question could still be answered by the method given:

    For example find the coeficient of x in the expansion of:

    (1+4x)^{1025} (3-2x)^{77}

    CB
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