Chapter 5.8, Problem 16PPS

To determine

To calculate: The rational zeroes of the function, $f\left(x\right)=15x^3+6x^2+x+90$ .

Answer to Problem 16PPS

The rational zeroes of the function $f\left(x\right)=15x^3+6x^2+x+90$ are $\begin{array}{l}\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 90\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{6}{5},\pm \frac{9}{5},\pm 18\\ \text{and}\pm \frac{1}{15},\pm \frac{2}{15},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{2}{3}\end{array}$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=15x^3+6x^2+x+90$

Formula used:

The rational zero theorem:

The polynomial function $p\left(x\right)$ with integral coefficients, then every rational zero of $p\left(x\right)=0$ is of the form $\frac{p}{q}$ , where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient.

Calculation:

Consider the polynomial function,

  $f\left(x\right)=15x^3+6x^2+x+90$

Apply the rational zero theorem,

  $f\left(x\right)=15x^3+6x^2+x+90=0$

If $\frac{p}{q}$ is a rational zero, then $p$ is a factor of the constant term $90$ and $q$ is a factor of the leading coefficient $15$ .

The possible values for $p$ is $\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 90$ and the possible values for $q$ is $\pm 1,\pm 3,\pm 5,\pm 15$ .

Now, the possible values of $\frac{p}{q}$ in simplest form,

  $\begin{array}{l}\frac{p}{q}=\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 90\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{6}{5},\pm \frac{9}{5},\pm 18\\ \text{and}\pm \frac{1}{15},\pm \frac{2}{15},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{2}{3}\end{array}$

Thus, the rational zeroes of the function $f\left(x\right)=15x^3+6x^2+x+90$ are $\begin{array}{l}\pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 90\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{5}{3},\pm \frac{10}{3},\pm \frac{1}{5},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{6}{5},\pm \frac{9}{5},\pm 18\\ \text{and}\pm \frac{1}{15},\pm \frac{2}{15},\pm \frac{2}{5},\pm \frac{3}{5},\pm \frac{2}{3}\end{array}$