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    x3+3x+k, put this into the form (ax2+bx+c)(x-1) and show k=2, the remainder is 6
    First I got the form (ax2+bx+c)(x-1)
    multiplied ax3+bx2+cx-ax2-bx-c
    grouped like terms ax3+(b-a)x2+(c-b)x-c
    Put original equation into that form 1x3+0x2+3x-(-k)
    1 is a, b-1=0 therefore b is also 1, c is -k, -k-1=3 therefore k should be 4 but its not, the answers (which do not show the working out) says its 2, which does satisfy the original equation if I substitute x=1.
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    Firstly, the remainder theorem tells you that f(1) = 0 so 1+3+k = 0
    So k = -4

    Are you sure that the question is correct?
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    Oh sorry it says the remainder is 6
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    (Original post by Primus2x)
    Oh sorry it says the remainder is 6
    So you have not given anything like the original question ... why not?

    Anyway 1+3+k=6 gives k=2 as required
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    (Original post by Primus2x)
    ax3+(b-a)x2+(c-b)x-c
    With your new information this will read

    ax^3 + (b-a)x^2 + (c-b)x - c + 6

    So this method also gives c=4 and k=2
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    I see it makes sense now so if f(x) gives a remainder when divided by (x-n) it means it is equal to the remainder when I substitute x=n into it?
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    (Original post by Primus2x)
    I see it makes sense now so if f(x) gives a remainder when divided by (x-n) it means it is equal to the remainder when I substitute x=n into it?
    That is indeed the Remainder Theorem
 
 
 
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