Chapter 8.9, Problem 28WE

a.

To determine

To find: the location ofreal zero of polynomial function between consecutive tenths.

Answer to Problem 28WE

The real zero lie between $P\left(2.4\right)=-.696\text{ and }P\left(2.5\right)=0.125$

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^3-2x^2-3$ .

Calculation: to find the real zero, first find the polynomial function values,

$x$	$P\left(x\right)$
$0$	$-3$
$2$	$-3$
$3$	$6$

From the above table it can be observed that the polynomial zero can be lie in between $2-3$ , so now find the polynomial function values between $2-3$ ,

$x$	$P\left(x\right)$
$2.3$	$-1.413$
$2.4$	$-0.696$
$2.5$	$0.125$

Thus, the real zero of polynomial must be between in $2.4-2.5$ .

b.

To determine

To find: the zero to the nearest hundredth using linear interpolation.

Answer to Problem 28WE

The real zero is $r=2.48$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=x^3-2x^2-3$ .

Calculation: to find the real zero, first find the polynomial function values,

$x$	$P\left(x\right)$
$0$	$-3$
$2$	$-3$
$3$	$6$

From the above table it can be observed that the polynomial zero can be lie in between $2-3$ , so now find the polynomial function values between $2-3$ ,

$x$	$P\left(x\right)$
$2.3$	$-1.413$
$2.4$	$-0.696$
$2.5$	$0.125$

Thus, the real zero of polynomial must be between in $2.4-2.5$ .

Let $r$ is the real zero of polynomial,

$x$	$P\left(x\right)$
$2.4$	$-0.696$
$r$	$0$
$2.5$	$0.125$

Now, to find the difference,

  $\begin{array}{l}\frac{c}{.1}=\frac{.696}{0.821}\\ \frac{c}{.1}=0.84\text{ [divide]}\\ c=0.084\text{ [multiply]}\end{array}$

Thus, the real zero of polynomial is,

  $\begin{array}{l}r=2.4+c\\ r=2.4+0.084\text{ [substitute]}\\ r=2.48\text{ [add]}\end{array}$