# Cubic function with integral coefficients.

• Nov 19th 2007, 03:49 PM
Jeavus
Cubic function with integral coefficients.
"A cubic function f(x) with integral coefficients has the following properties: f(3/2) = 0, (x-2) is a factor of f(x), and f(4) = 50. Determine f(x)."

How do I solve this?

The two factors are (x - 3/2) and (x - 2), right?

I don't know what to do from there.
• Nov 19th 2007, 04:24 PM
ThePerfectHacker
Quote:

Originally Posted by Jeavus
"A cubic function f(x) with integral coefficients has the following properties: f(3/2) = 0, (x-2) is a factor of f(x), and f(4) = 50. Determine f(x)."

How do I solve this?

The two factors are (x - 3/2) and (x - 2), right?

I don't know what to do from there.

So,
$\displaystyle f(x) = A(x-3/2)(x-2)(x-c)$
And $\displaystyle f(4)=50$ so,
$\displaystyle 50 = A(4-3/2)(4-2)(4-c)$
It seems you need another condition.
• Nov 19th 2007, 04:34 PM
Soroban
Hello, Jeavus!

Quote:

A cubic function f(x) with integral coefficients has the following properties:
. . $\displaystyle f\left(\frac{3}{2}\right) \:=\: 0,\;\;(x-2)$ is a factor of $\displaystyle f(x)$, and $\displaystyle f(4) \:= \:50$
Determine $\displaystyle f(x)$.

Two of the factors are: .$\displaystyle (x - 2)$ and $\displaystyle (2x-3)$

Since the function is a cubic, there is a third factor: .$\displaystyle (x - a)$

Hence: .$\displaystyle f(x) \;\;=\;\;(x-2)(2x-3)(x-a) \;\;=\;\;2x^3 - (2a+7)x^2 + (7a+6)x - 6a$

Since $\displaystyle f(4) = 50$, we have: .$\displaystyle 2\!\cdot\!4^3 - (2a+7)\!\cdot\!4^2 + (7a+6)\!\cdot\!4 - 6a\;=\;50$

. . which simplifies to: .$\displaystyle a \:=\:-1$

Therefore, the cubic is: .$\displaystyle {\color{blue}f(x)\;=\;2x^3 - 5x^2 - x + 6}$