POLYNOMIALS
posted by Aoi .
The cubic polynomial f(x) is such that the coefficient of x^3 is 1. and the roots of the equation f(x) = 0 are 1, 2 and k. Given that f(x)has a remainder of 8 when divided by (x3), find the value of k.
okay, this is what i did:
x^3 + bx^2 + cx + d = (x1)(x2)(xk)
f(3)=8
x^3 + bx^2 +cx +d = (x1)(x^2kx2x+2k)
and im stuck...
is my interpretation of the question correct? when the questions states that the coefficient of x^3 is 1, it means the polynomial is something like x^3 + ax^2 + bx + c right???

don't expand it.
let f(x) = (x1)(x2)(xk)
it should be clear that if we would expand the right side, the first term would be x^3
given f(3) = 8 , so ...
(2)(1)(3k) = 9
2(3k) = 8
6 + 3k = 8
3k = 15
k = 5 
There's a bit of calculation mistakes there, but thanks, i know how to do it now!!!! XD