Chapter 6.8, Problem 17PPS

To determine

To calculate: The rational zeroes of the function, $f\left(x\right)=16x^4-5x^2+128$ .

Answer to Problem 17PPS

The rational zeroes of the function $f\left(x\right)=16x^4-5x^2+128$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8}\text{and}\pm \frac{1}{16}$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=16x^4-5x^2+128$

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,

  $f\left(x\right)=16x^4-5x^2+128$

Apply the rational zero theorem,

  $f\left(x\right)=16x^4-5x^2+128=0$

If $\frac{p}{q}$ is a rational zero, then $p$ is a factor of the constant term $128$ and $q$ is a factor of the leading coefficient $16$ .

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

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

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8},\pm \frac{1}{16}$

Thus, the rational zeroes of the function $f\left(x\right)=16x^4-5x^2+128$ are $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm 32,\pm 64,\pm 128,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{8}\text{and}\pm \frac{1}{16}$