Chapter 6.8, Problem 13PPS

To determine

To calculate: The rational zeroes of the function, $f\left(x\right)=3x^6-4x^4-x^2-35$ .

Answer to Problem 13PPS

The rational zeroes of the function $f\left(x\right)=3x^6-4x^4-x^2-35$ are $\pm 1,\pm 5,\pm 7,\pm 35,\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{7}{3}\text{and}\pm \frac{35}{3}$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=3x^6-4x^4-x^2-35$

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,

  $f\left(x\right)=3x^6-4x^4-x^2-35$

Apply the rational zero theorem,

  $f\left(x\right)=3x^6-4x^4-x^2-35=0$

If $\frac{p}{q}$ is a rational zero, then $p$ is a factor of the constant term $-35$ and $q$ is a factor of the leading coefficient $3$ .

The possible values for $p$ is $\pm 1,\pm 5,\pm 7,\pm 35$ and the possible values for $q$ is $\pm 1,\pm 3$ .

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

  $\frac{p}{q}=\pm 1,\pm 5,\pm 7,\pm 35,\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{7}{3},\pm \frac{35}{3}$

Thus, the rational zeroes of the function $f\left(x\right)=2x^3+5x^2-8x-10=0$ are $\pm 1,\pm 5,\pm 7,\pm 35,\pm \frac{1}{3},\pm \frac{5}{3},\pm \frac{7}{3}\text{and}\pm \frac{35}{3}$