Chapter 2.6, Problem 14E

Solution

To find: all real zeros of the function.

The three zeros of the function are $-8,-3,3$ .

Given information: 

The function $h\left(x\right)=x^3+8x^2-9x-72$ .

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Consider the function

  $h\left(x\right)=x^3+8x^2-9x-72$ .

Then

Factors of constant term $72$ :

  $p=1,2,3,4,6,8,9,12,18,24,36,72$

Leading coefficient factors $1$ :

  $q=1$

All possible combinations of $\pm \frac{p}{q}$ .

Some or none may be zeros of the function.

Simplify and eliminate duplicate entries.

  $\pm \frac{p}{q}=\pm \left(1,2,3,4,6,8,9,12,18,24,36,72\right)$

Test the possible rational zeros.

I poly horner scheme $[x=1]\left\{x^3+8x^2-9x-72\right\}$

  $\backslash$ polyhornerscheme $[x=2]\left\{x^3+8x^2-9x-72\right\}$

I polyhornerscheme $[x=3][x^3+8x^2-9x-72\}$

  $3$ is a zero, so $\left(x-3\right)$ is a factor.

Using the results of the division.

  $\begin{array}{c}h\left(x\right)=x^3+8x^2-9x-72\\ =\left(x-3\right)\left(x^2+11x+24\right)\end{array}$ and factor the remaining part

  $\begin{array}{c}h\left(x\right)=\left(x-3\right)\left(x+3\right)\left(x+8\right)\\ x=-8,-3,3\end{array}$