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    (2+x)^5 + (2-x)^5 = A + Bx^2 + Cx^4

    the equals sign has 3 lines (what exactly does that mean anyways)

    I know the answer, but using my method I can only get A, what is the correct method to get all 3 constants?

    Thanks,

    W.
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    I'd do a binomial expansion of both brackets and then compare terms once simplified.


    The three lines on the equals sign just means 'equivalent to'. Not much in it really.
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    (Original post by wonderboy)
    (2+x)^5 + (2-x)^5 = A + Bx^2 + Cx^4

    the equals sign has 3 lines (what exactly does that mean anyways)

    I know the answer, but using my method I can only get A, what is the correct method to get all 3 constants?

    Thanks,

    W.
    the three lines means equivalent to, or identical to, in other words, regardless of the value of x, there will always be some A, B and C where it will satisfy that equation!

    If I were you, I'd expand the brackets out, then simplfy; so... let me think...
    coefficents will be 1 5 10 10 5 1
    giving you :-
    (2+x)^5 = 2^5 + 5(2^4) x + 10(2^3)x^2 + 10(2^2)x^3 + 5(2)x^4 + x^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5

    and (2-x)^5 = 2^5 + 5(2^4) -x + 10(2^3)(-x)^2 + 10(2^2)(-x)^3 + 5(2)(-x)^4 + (-x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5

    therefore (2+x)^5 + (2-x)^5 = 64 + 160x^2 + 20x^4
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    The three lines means that it is an identity and not an equation.
    An identity is true for all values of x, if you try and solve an identity (which you shouldnt !!) like an equation then you will end up with 0 = 0.

    An equation is true for a finite number of values of x.
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    (Original post by Katie Heskins)
    the three lines means equivalent to, or identical to, in other words, regardless of the value of x, there will always be some A, B and C where it will satisfy that equation!

    If I were you, I'd expand the brackets out, then simplfy; so... let me think...
    coefficents will be 1 5 10 10 5 1
    giving you :-
    (2+x)^5 = 2^5 + 5(2^4) x + 10(2^3)x^2 + 10(2^2)x^3 + 5(2)x^4 + x^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5

    and (2-x)^5 = 2^5 + 5(2^4) -x + 10(2^3)(-x)^2 + 10(2^2)(-x)^3 + 5(2)(-x)^4 + (-x)^5 = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5

    therefore (2+x)^5 + (2-x)^5 = 64 + 160x^2 + 20x^4
    Thanks, I won't say how I tried to do it
    W.
 
 
 
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