Cubic function with integral coefficients.

**Jeavus** · November 19th 2007, 03:49 PM

"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.

**ThePerfectHacker** · November 19th 2007, 04:24 PM

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,
$f(x) = A(x-3/2)(x-2)(x-c)$
And $f(4)=50$ so,
$50 = A(4-3/2)(4-2)(4-c)$
It seems you need another condition.

**Soroban** · November 19th 2007, 04:34 PM

Hello, Jeavus!

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

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

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

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

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

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

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

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