Chapter 5.8, Problem 26PPS

To determine

To calculate: The rational zeroes of the function, $f\left(x\right)=81x^4-256$ .

Answer to Problem 26PPS

The rational zeroes of the function $f\left(x\right)=81x^4-256$ are

  $\begin{array}{l}\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256,\\ \pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3},\pm \frac{16}{3},\pm \frac{32}{3},\pm \frac{64}{3},\pm \frac{128}{3},\pm \frac{256}{3},\\ \pm \frac{1}{9},\pm \frac{2}{9},\pm \frac{4}{9},\pm \frac{8}{9},\pm \frac{16}{9},\pm \frac{32}{9},\pm \frac{64}{9},\pm \frac{128}{9},\pm \frac{256}{9},\\ \pm \frac{1}{27},\pm \frac{2}{27},\pm \frac{4}{27},\pm \frac{8}{27},\pm \frac{16}{27},\pm \frac{32}{27},\pm \frac{64}{27},\pm \frac{128}{27},\pm \frac{256}{27},\\ \text{and}\\ \pm \frac{1}{81},\pm \frac{2}{81},\pm \frac{4}{81},\pm \frac{8}{81},\pm \frac{16}{81},\pm \frac{32}{81},\pm \frac{64}{81},\pm \frac{128}{81},\pm \frac{256}{81}\end{array}$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=81x^4-256$

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,

  $f\left(x\right)=81x^4-256$

Apply the rational zero theorem,

  $f\left(x\right)=81x^4-256=0$

If $\frac{p}{q}$ is a rational zero, then $p$ is a factor of the constant term $-256$ and $q$ is a factor of the leading coefficient $81$ .

The possible values for $p$ is $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256$ and the possible values for $q$ is $\pm 1,\pm 3,\pm 9,\pm 27,\pm 81$ .

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

  $\begin{array}{l}\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256,\\ \pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3},\pm \frac{16}{3},\pm \frac{32}{3},\pm \frac{64}{3},\pm \frac{128}{3},\pm \frac{256}{3},\\ \pm \frac{1}{9},\pm \frac{2}{9},\pm \frac{4}{9},\pm \frac{8}{9},\pm \frac{16}{9},\pm \frac{32}{9},\pm \frac{64}{9},\pm \frac{128}{9},\pm \frac{256}{9},\\ \pm \frac{1}{27},\pm \frac{2}{27},\pm \frac{4}{27},\pm \frac{8}{27},\pm \frac{16}{27},\pm \frac{32}{27},\pm \frac{64}{27},\pm \frac{128}{27},\pm \frac{256}{27},\\ \text{and}\\ \pm \frac{1}{81},\pm \frac{2}{81},\pm \frac{4}{81},\pm \frac{8}{81},\pm \frac{16}{81},\pm \frac{32}{81},\pm \frac{64}{81},\pm \frac{128}{81},\pm \frac{256}{81}\end{array}$

Thus, the rational zeroes of the function $f\left(x\right)=81x^4-256$ are $\begin{array}{l}\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm 256,\\ \pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3},\pm \frac{16}{3},\pm \frac{32}{3},\pm \frac{64}{3},\pm \frac{128}{3},\pm \frac{256}{3},\\ \pm \frac{1}{9},\pm \frac{2}{9},\pm \frac{4}{9},\pm \frac{8}{9},\pm \frac{16}{9},\pm \frac{32}{9},\pm \frac{64}{9},\pm \frac{128}{9},\pm \frac{256}{9},\\ \pm \frac{1}{27},\pm \frac{2}{27},\pm \frac{4}{27},\pm \frac{8}{27},\pm \frac{16}{27},\pm \frac{32}{27},\pm \frac{64}{27},\pm \frac{128}{27},\pm \frac{256}{27},\\ \text{and}\\ \pm \frac{1}{81},\pm \frac{2}{81},\pm \frac{4}{81},\pm \frac{8}{81},\pm \frac{16}{81},\pm \frac{32}{81},\pm \frac{64}{81},\pm \frac{128}{81},\pm \frac{256}{81}\end{array}$