# cubic polynomial

Printable View

• Jan 1st 2011, 11:20 PM
caramelcake
cubic polynomial
Quote:

The cubic polynomial \$\displaystyle f(x) \$is such that the coefficient of \$\displaystyle x^3\$ is \$\displaystyle 1\$ and the roots of \$\displaystyle f(x) = 0\$ are \$\displaystyle k\$, \$\displaystyle 2k\$ and \$\displaystyle (1 - k)\$, where \$\displaystyle k > 0\$. It is given that \$\displaystyle f(x)\$ has a remainder of \$\displaystyle 30\$ when divided by (x - 1).
Show that \$\displaystyle 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 \$\displaystyle f(1) = 30\$
Does 'root' here mean factor?
Would that mean that \$\displaystyle f(x) = (k)(2k)(1 - k)\$? But that doesn't make much sense to me because I don't see how \$\displaystyle 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.
• Jan 1st 2011, 11:43 PM
CaptainBlack
Quote:

Originally Posted by caramelcake
I do not quite understand the question so I don't know where to start from.
I only know that \$\displaystyle f(1) = 30\$
Does 'root' here mean factor?
Would that mean that \$\displaystyle f(x) = (k)(2k)(1 - k)\$? But that doesn't make much sense to me because I don't see how \$\displaystyle 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.

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

CB
• Jan 2nd 2011, 12:20 AM
caramelcake
Thank you! The question is surprisingly simple now :)