Chapter 2.5, Problem 27E

To determine

Tofind:the rational zeroes of the function.

Answer to Problem 27E

Therational zeroes of the functions are at $x=3,-2,\pm \frac{2}{3}.$

Explanation of Solution

Given:

  ${\text{9x}}^{\text{4}}{\text{-9x}}^{\text{3}}{\text{-58x}}^{\text{2}}\text{+4x+24}$

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:

  ${\text{9x}}^{\text{4}}{\text{-9x}}^{\text{3}}{\text{-58x}}^{\text{2}}\text{+4x+24}$

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.

  ${\text{9x}}^{\text{3}}\left(\text{x-3}\right){\text{+27x}}^{\text{3}}{\text{-9x}}^{\text{3}}{\text{-58x}}^{\text{2}}\text{+4x+24}$

  ${\text{9x}}^{\text{3}}\left(\text{x-3}\right){\text{+18x}}^{\text{2}}\left(\text{x-3}\right){\text{-4x}}^{\text{2}}\text{+4x+24}$

  ${\text{9x}}^{\text{3}}\left(\text{x-3}\right){\text{+18x}}^{\text{2}}\left(\text{x-3}\right)\text{-4x}\left(\text{x-3}\right)\text{-8}\left(\text{x-3}\right)$

  $\left(\text{x-3}\right)\left({\text{9x}}^{\text{3}}{\text{+18x}}^{\text{2}}\text{-4x-8}\right)$ .

  $\left(\text{x-3}\right)\left({\text{9x}}^{\text{2}}\left(\text{x+2}\right){\text{-18x}}^{\text{2}}{\text{+18x}}^{\text{2}}\text{-4}\left(\text{x+2}\right)\right)$

  $\left(\text{x-3}\right)\left(\text{x+2}\right)\left({\text{9x}}^{\text{2}}\text{-4}\right)$

  $x=3,-2,\pm \frac{2}{3}.$

Hence the rational zeroes of the functions are at $x=3,-2,\pm \frac{2}{3}.$