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    In the expansion of (1 - 3x)(1 + kx)^9, the coefficient of x^6 is zero. Find the non-zero value of k.

    Could somebody please explain this question?
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    (Original post by ThugzMansion7)
    In the expansion of (1 - 3x)(1 + kx)^9, the coefficient of x^6 is zero. Find the non-zero value of k.

    Could somebody please explain this question?
    You need to choose the most suitable terms to multiply out, in terms of k.
    Clearly we want the
    coefficients of x^5 and x^6 from (1+kx) as we multiply by both -3x and 1 to obtain terms with x^6.
    Once you've found the coefficient you can set it as 0 and hence find k. (but remember obviously k won't be 0).
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    (Original post by Gaz031)
    You need to choose the most suitable terms to multiply out, in terms of k.
    Clearly we want the
    coefficients of x^5 and x^6 from (1+kx) as we multiply by both -3x and 1 to obtain terms with x^6.
    Once you've found the coefficient you can set it as 0 and hence find k. (but remember obviously k won't be 0).
    why do i take away the x^6 term from the X^5 term?
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    (Original post by ThugzMansion7)
    why do i take away the x^6 term from the X^5 term?
    I never said that.
    You need both the x^5 and x^6 from the second binomial to ensure you can the correct answer when multiplying by (1-3x)
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    but if you work through the answer, you have to take away one from the other. It's in my revise for p2 book (page 17 ex. 4)
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    (Original post by ThugzMansion7)
    but if you work through the answer, you have to take away one from the other. It's in my revise for p2 book (page 17 ex. 4)
    We need:
    (1-3x)(...126k^5x^5+84k^6x^6)
    The coefficients of x^6 is 84k^6-3(126k^5)
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    do you take that away from the coeff of x^5?
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    (Original post by ThugzMansion7)
    do you take that away from the coeff of x^5?
    You get rid of the coefficient of x^5 after the multiplication...
    (1-3x)(...126k^5x^5+84k^6x^6)
    =126k^5x^5+84k^6x^6-3x(126k^5x^5)-3x(84k^6x^6)
    so we want x^6[84k^6-3(126k^5)]
    coefficient =0
    84k^6-3(126)k^5=0
    k not 0 otherwise we wouldn't have most of our expansion:
    k=3(126/84).
 
 
 
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