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 (x-3), find the value of k.

Question

okay, this is what i did:

-x^3 + bx^2 + cx + d = (x-1)(x-2)(x-k)
f(3)=8
-x^3 + bx^2 +cx +d = (x-1)(x^2-kx-2x+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???

Answer 1

don't expand it.
let f(x) = -(x-1)(x-2)(x-k)
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)(3-k) = 9
-2(3-k) = 8
-6 + 3k = 8
3k = 15
k = 5

Answer 2

There's a bit of calculation mistakes there, but thanks, i know how to do it now!!!! XD

Answer 3

let f(x)= -(x-1)(x-2)(x-k)
given that f(3)=8,
-(2)(1)(3-k)=8
-2(3-k)=8
-6+2k=8
2k=14
k=7

hope this helped