Chapter SH, Problem 5.18EP

To determine

Find the zeros of the given function.

Answer to Problem 5.18EP

The function has zeros at $-2,1$ and $5$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^3-4x^2-7x+10$ .

Calculation:

Since $f\left(x\right)$ has degree $3$ , the function has $3$ zeros.

Find the type of zeros.

Examine the number of sign changes for $f\left(x\right)$ and $f\left(-x\right)$ .

  $\begin{array}{l}f\left(x\right)=x^3-4x^2-7x+10\\ \text{yes}\text{no}\text{yes}\end{array}$

Because there are two sign changes for the coefficients of $f\left(x\right)$ , the function has two or zero positive real zeros.

  $\begin{array}{l}f\left(-x\right)={\left(-x\right)}^3-4{\left(-x\right)}^2-7\left(-x\right)+10\\ =-x^3-4x^2+7x+10\\ \text{no}\text{yes}\text{no}\end{array}$

Because there is one sign changes for the coefficients of $f\left(-x\right)$ , the function has one or zero negative real zeros.

Thus, $f\left(x\right)$ has three real zeros or three imaginary zeros.

Determine the real zeros. List some possible values, and then use synthetic substitution to evaluate $f\left(x\right)$ for real values of $x$ .

$x$	$1$	$-4$	$-7$	$10$
$-2$	$1$	$-6$	$5$	$0$
$-1$	$1$	$-5$	$2$	$8$
$0$	$1$	$-4$	$-7$	$10$
$1$	$1$	$-3$	$-10$	$0$
$2$	$1$	$-2$	$-11$	$-12$

From the table, zero occurs at $x=-2$ and $x=1$ . Since there are one negative real zero and two positive real zeros.

Use synthetic substitution with depressed polynomial $f\left(x\right)=x^2-3x-10$ to find second positive real zero.

$x$	$1$	$-4$	$-7$	$10$
$3$	$1$	$-1$	$-10$	$-20$
$4$	$1$	$0$	$-7$	$10$
$5$	$1$	$1$	$-2$	$0$

Hence the function haszeros at $-2,1$ and $5$ .