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ok, this might be a bit torturous, but if anyone is patient enough and help me with my working, i'd REALLY APPRECIATE IT!!!!

The cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k, and k^2. IT is given that f(x) has a remainder of 7 when divided by x-2.
Show that k^3 - 2k^2 - 2k - 3 = 0 __________________________________________________________________________

this is what i did, it might be a stupid or completely wrong way of doing..... please pardon my stupidity

x^3 + ax^2 + bx^2 + c = (x-1)(x-k)(x-k^2)

(x-1)(x-k)(x-k^2)= x^3 + (-k-1-k^2)x^2 + (-k^3+k^2+k)x -k^3

so... (-k-1-k^2)= a
(-k^3 + k^2 + k)=b
-k^3 =c
4(-k^2-k-1)+2(-k^3+k^2+k)-k^3= -1
-3k^3-2k^2-2k-3= 0

!!??? which is different from what i;m supposed to show!?!?!?
I honestly don't know what im doing.........


    ok... i know how to do this question already! :D


    For the previous part, this was how i got the answer:
    f(x) = (x-1)(x-k)(x-k^2)
    = Q(x)(2-x)+7
    Let x=2, so (2-k)(2-k^2)=7, and i get the answer,

    But there is another part to this question which i don't understand...
    please help:D
    it says: Hence, find a value of k and show that there are no other real values of k which satisfy this equation.

    How do i do this??


    k^3 - 2k^2 - 2k - 3 = 0
    after a few quick trial & error guesses
    I found k=3 to work
    so by division I got
    k^3 - 2k^2 - 2k - 3 = 0
    (k-3)(k^2 + k + 1) = 0

    so any other solutions for k must come from
    k^2 + k + 1=0
    Using the formula it can be quickly shown that there is no "real" solution to this

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