Posted by **Aoi** on Monday, June 15, 2009 at 10:19am.

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.

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???

- POLYNOMIALS -
**Reiny**, Monday, June 15, 2009 at 1:39pm
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

- POLYNOMIALS -
**Aoi**, Monday, June 15, 2009 at 10:11pm
There's a bit of calculation mistakes there, but thanks, i know how to do it now!!!! XD

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