Chapter SH, Problem 5.19EP

To determine

Find the zeros of the given function.

Answer to Problem 5.19EP

The function has zeros at $-3,3,-i$ and $i$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=x^4-8x^2-9$ .

Calculation:

Since $f\left(x\right)$ has degree $4$ , the function has $4$ zeros.

Find the type of zeros.

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

  $\begin{array}{l}f\left(x\right)=x^4-8x^2-9\\ \text{yes}\text{no}\end{array}$

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

  $\begin{array}{l}f\left(-x\right)={\left(-x\right)}^4-8{\left(-x\right)}^2-9\\ =x^4-8x^2-9\\ \text{yes}\text{no}\end{array}$

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 two real zerosand two imaginary zeros or four 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 $x$ .

$x$	$1$	$0$	$-8$	$0$	$-9$
$-3$	$1$	$-3$	$1$	$-3$	$0$
$-2$	$1$	$-2$	$-4$	$8$	$-25$
$-1$	$1$	$-1$	$-7$	$7$	$-16$
$0$	$1$	$0$	$-8$	$0$	$-9$
$1$	$1$	$1$	$-7$	$-7$	$-16$
$2$	$1$	$2$	$-4$	$-8$	$-25$
$3$	$1$	$3$	$1$	$3$	$0$

From the table, zero occurs at $x=-3$ and $x=3$ . Since there are one negative real zero and one positive real zeros.

Use synthetic substitution with depressed polynomial $f\left(x\right)=x^3+3x^2+x+3$ to find two imaginary zeros.

Substitute $f\left(x\right)=0$ .

  $x^3+3x^2+x+3=0$

Use factorization method.

  $\begin{array}{l}{x}^3+3x^2+x+3=0\\ \Rightarrow x^2\left(x+3\right)+\left(x+3\right)=0\\ \Rightarrow \left(x+3\right)\left(x^2+1\right)=0\\ \Rightarrow x^2+1=0\\ \Rightarrow x^2=-1\\ \Rightarrow x=\pm \sqrt{-1}\\ i^2=-1\\ \Rightarrow x=\pm \sqrt{i^2}\\ \Rightarrow x=\pm i\end{array}$

Hence the function haszeros at $-3,3,-i$ and $i$ .