cubic polynomial

**caramelcake** · January 1st 2011, 11:20 PM

The cubic polynomial $f(x)$ is such that the coefficient of $x^3$ is $1$ and the roots of $f(x) = 0$ are $k$ , $2k$ and $(1 - k)$ , where $k > 0$ . It is given that $f(x)$ has a remainder of $30$ when divided by (x - 1).
Show that $2k^3 - 3k^2 + k - 30 = 0$

I do not quite understand the question so I don't know where to start from.
I only know that $f(1) = 30$
Does 'root' here mean factor?
Would that mean that $f(x) = (k)(2k)(1 - k)$ ? But that doesn't make much sense to me because I don't see how $2k^2 - 2k^3$ has any connection with anything.

I'm sorry for the silly question and any help would be really really appreciated. Thank you in advance.

**CaptainBlack** · January 1st 2011, 11:43 PM

Originally Posted by caramelcake

I do not quite understand the question so I don't know where to start from.
I only know that $f(1) = 30$
Does 'root' here mean factor?
Would that mean that $f(x) = (k)(2k)(1 - k)$ ? But that doesn't make much sense to me because I don't see how $2k^2 - 2k^3$ has any connection with anything.

I'm sorry for the silly question and any help would be really really appreciated. Thank you in advance.

$f(x)=(x-k)(x-2k)(x-(1-k))$

CB

**caramelcake** · January 2nd 2011, 12:20 AM

Thank you! The question is surprisingly simple now

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