Chapter 4.5, Problem 34E

To determine

To find: all real zeros of the given function.

Answer to Problem 34E

The real zeros are $-2,4$ , and $6$ .

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=x^3-28x-48$

Calculation:

Consider the function.

  $g\left(x\right)=x^3-28x-48$

The above function is of degree 3, so it has 3 zeros.

Here, the leading coefficient is $1$ and constant term is $-48$ .

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 6,\pm 8,\pm 12,\pm 24\text{and}\pm 48$ .

Test possible zeros using synthetic division until a solution is found.

For $x=-2$

  $\begin{array}{ccccc}-2& 1& 0& -28& -48\\ & & -2& 4& 48\\ & 1& -2& -24& 0\end{array}$

Therefore, $x+2$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}g\left(x\right)=x^3-28x-48\\ =\left(x+2\right)\left(x^2-2x-24\right)\\ =\left(x+2\right)\left(x-6\right)\left(x-4\right)\end{array}$

Solve further for $g\left(x\right)=0$ .

  $\begin{array}{c}\left(x+2\right)\left(x-6\right)\left(x-4\right)=0\\ x=-2,4,6\end{array}$

Therefore, all real zeros are $-2,4$ , and $6$ .