Chapter 4.5, Problem 33E

To determine

To find: all real zeros of the given function.

Answer to Problem 33E

The real zeros are $-5,4$ , and $3$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^3-2x^2-23x+60$

Calculation:

Consider the function.

  $f\left(x\right)=x^3-2x^2-23x+60$

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

Here, the leading coefficient is $1$ and constant term is $60$ .

The possible rational zeros are $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30\text{and}\pm 60$ .

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

For $x=3$

  $\begin{array}{ccccc}3& 1& -2& -23& 60\\ & & 3& 3& -60\\ & 1& 1& -20& 0\end{array}$

Therefore, $x-3$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}f\left(x\right)=x^3-2x^2-23x+60\\ =\left(x-3\right)\left(x^2+x-20\right)\\ =\left(x+5\right)\left(x-4\right)\left(x-3\right)\end{array}$

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

  $\begin{array}{c}\left(x+5\right)\left(x-4\right)\left(x-3\right)=0\\ x=-5,4,3\end{array}$

Therefore, all real zeros are $-5,4$ , and $3$ .