Chapter 1.2, Problem 9E

To determine

To find: An expression for a cubic function which satisfies the given conditions.

Answer to Problem 9E

The cubic function is $3\left(x^3-x^2-2x\right)$.

Explanation of Solution

Given:

The given conditions are $f\left(1\right)=6,f\left(-1\right)=f\left(0\right)=f\left(2\right)=0$.

Observation:

The number a is said to be a zero of an equation when $f\left(a\right)=0$, where $x-a$ is said to be a factor of the equation.

Since $f\left(-1\right)=f\left(0\right)=f\left(2\right)=0$, −1, 0, 2 are the roots of the linear equation and $\left(x+1\right)\left(x-0\right)\text{ and }\left(x-2\right)$ are the factors of the equation.

Thus, $f\left(x\right)=a\left(x+1\right)\left(x\right)\left(x-2\right)$.

Since $f\left(1\right)=6$, the point (1, 6) satisfies the given condition.

Substitute 1 for x and 6 for $f\left(x\right)$ and obtain the value of a,

$\begin{array}{l}6=a\left(2\right)1\left(-1\right)\\ -2a=6\\ a=-3\end{array}$

Thus, the value of $a=-3$.

Therefore, the cubic equation that satisfies the conditions is $f\left(x\right)=-3x\left(x+1\right)\left(x-2\right)$.

That is, the cubic function is $f\left(x\right)=3\left(x^3-x^2-2x\right)$.