POLYNOMIALS
posted by Aoi .
ok, this might be a bit torturous, but if anyone is patient enough and help me with my working, i'd REALLY APPRECIATE IT!!!!
Question:
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 x2.
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 = (x1)(xk)(xk^2)
f(2)=7
8+4a+2b+c=7
4a+2b+c=1
(x1)(xk)(xk^2)= x^3 + (k1k^2)x^2 + (k^3+k^2+k)x k^3
so... (k1k^2)= a
(k^3 + k^2 + k)=b
k^3 =c
4(k^2k1)+2(k^3+k^2+k)k^3= 1
3k^32k^22k3= 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) = (x1)(xk)(xk^2)
= Q(x)(2x)+7
Let x=2, so (2k)(2k^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
(k3)(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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