Chapter 2.5, Problem 29E

To determine

Tofind:the real solution of the polynomial function.

Answer to Problem 29E

Thereal solution of the given equation is $\text{z=1,-2}$ .

Explanation of Solution

Given:

  ${\text{z}}^{\text{4}}{\text{+z}}^{\text{3}}{\text{+z}}^{\text{2}}\text{+3z-6}$

Concept used:

The factor theorem:

  $\left(\text{x-r}\right)$ is the factor of polynomial if and only if r is the root of the $\text{p}\left(\text{x}\right)$ .

By taking the factor common out thus generate the new equation in the form:

  $\left(\text{x-r}\right)\text{E}$ .

That equation $\text{E}$ has the degree lesser than before and again taking the common factor from the $\text{E}$ until the remainders equal to 0.

Calculation:

  ${\text{z}}^{\text{4}}{\text{+z}}^{\text{3}}{\text{+z}}^{\text{2}}\text{+3z-6}$

By putting $x=3$ the function equal to 0.

Therefore $\left(\text{x-3}\right)$ is one factor of the given function. Other factor can be known by dividing the factor by the function.

  ${\text{z}}^{\text{4}}{\text{+z}}^{\text{3}}{\text{+z}}^{\text{2}}\text{+3z-6}$

  ${\text{z}}^{\text{3}}\left(\text{z-1}\right){\text{+2z}}^{\text{3}}{\text{+z}}^{\text{2}}\text{+3z-6}$

  ${\text{z}}^{\text{3}}\left(\text{z-1}\right){\text{+2z}}^{\text{2}}\left(\text{z-1}\right){\text{+3z}}^{\text{2}}\text{+3z-6}$

  ${\text{z}}^{\text{3}}\left(\text{z-1}\right){\text{+2z}}^{\text{2}}\left(\text{z-1}\right)\text{+3z}\left(\text{z-1}\right)\text{+6}\left(\text{z-1}\right)$ .

  $\left(\text{z-1}\right)\left({\text{z}}^{\text{3}}{\text{+2z}}^{\text{2}}\text{+3z+6}\right)$

  $\left(\text{z-1}\right)\left({\text{z}}^{\text{2}}\left(\text{z+2}\right)\text{+3}\left(\text{z+2}\right)\right)$

  $\left(\text{z-1}\right)\left(\text{z+2}\right)\left({\text{z}}^{\text{2}}\text{+3}\right)$

Since the $\left({\text{z}}^{\text{2}}\text{+3}\right)$ is not real it will be imaginary number $\text{z=±3i}$ which is not real

Hence the real solution of the given equation is $\text{z=1,-2}$ .