Chapter 6.8, Problem 23PT

To determine

To find: the possible number of positive, negative, imaginary real zeroes of the given function.

Answer to Problem 23PT

The table for the possible combination of real and imaginary zeroes is:

Positive real zeroes	Negative Real zeroes	Imaginary real zeroes	Total zeroes
$0$	$3$	$0$	$3$
$0$	$3$	$0$	$3$

Explanation of Solution

Given:

  $p\left(x\right)=x^3+2x^2+4x+8$ .

Concept used:

Descartes rule of sign:

Look at the change in the sign of the positive to negative or negative to positive.

The number of positive real zeroesis either equal to number of sign change of $f\left(x\right)$ or it could be less than positive even integer.

For negative real zeroes the $f\left(-x\right)$ is considered.

Look at the change in the sign of the positive to negative or negative to positive.

The number of negative real zeroes is either equal to number of sign change of $f\left(-x\right)$ or it could be less than positive even integer.

Sum Positive real zeroes, negative real zeroes and imaginary zeroes equals to the total zeroes of the polynomial through this the imaginary zeroes can be calculated.

Calculation:

According to the given the polynomial equation is:

  $\begin{array}{l}p\left(x\right)=x^3+2x^2+4x+8.\\ p\left(-x\right)=-x^3+2x^2-4x+8.\end{array}$

The total zeroes here is $3$ since, the highest degree of the polynomial is cubic polynomial.

Solving through Descartes rule of sign as:

The number of positive real zeroes is either equal to number of sign change of $f\left(x\right)$ or it could be less than positive even integer.

Similarly,

The number of negative real zeroes is either equal to number of sign change of $f\left(-x\right)$ or it could be less than positive even integer.

There is one sign change in $f\left(x\right)$ so, there is one possible positive real zero.

There is two sign change in $f\left(-x\right)$ so, there is two possible negative real zeroes.

Hence, the table for the possible combination of real and imaginary zeroes is:

Positive real zeroes	Negative Real zeroes	Imaginary real zeroes	Total zeroes
$0$	$3$	$0$	$3$
$0$	$3$	$0$	$3$