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Math Help - What does this mean in my exam practice paper?

  1. #1
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    What does this mean in my exam practice paper?

    Find the independant of x in the expansion of [x-(2/x)]^4

    Thanks, Rob
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  2. #2
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    That means find the constant term, the term without x. In binomial expansions, this is usually the term with x^0 (x to the power zero)
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  3. #3
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    I'm not sure that I quite understand that. Surely with the binomial expansion shown above, all of the expanded terms world have an x somewhere.
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  4. #4
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    Note:  \left(x-\frac{2}{x}\right)^4 can be written as  (x-2(x^{-1}))^4

    Just looking at the x terms (ignoring coefficients):

     (x^4)(x^0) + (x^3)(x^{-1}) + (x^2)(x^{-2}) + (x^1)(x^{-3}) + (x^0)(x^{-4})

    The third term cancels out.
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  5. #5
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    Hello, Rob!

    Don't argue about it . . . just expand it!


    Find the constant term in the expansion of . \left(x-\frac{2}{x}\right)^4

    \left(x-\frac{2}{x}\right)^4 \;\;=\;\;(x)^4 - \:4(x)^3\!\left(\frac{2}{x}\right) + \:6(x)^2\!\left(\frac{2}{x}\right)^2 - \:4(x)^1\!\left(\frac{2}{x}\right)^3 + \left(\frac{2}{x}\right)^4

    . . . . . . . = \;\;x^4 \:- \;4x^3\!\cdot\!\frac{2}{x} \:+ \;6x^2\!\cdot\!\frac{4}{x^2} \:- \;4x\!\cdot\!\frac{8}{x^3} \:+ \;\frac{16}{x^4}

    . . . . . . . = \;\;x^4 \:-\: 8x^2 \:+\: \underbrace{{\color{blue}24}}_{\uparrow} \:-\: \frac{32}{x^2} \:+\: \frac{16}{x^4}

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