Chapter 6.8, Problem 25PT

To determine

To calculate: The rational zeroes of the function, $f\left(x\right)=2x^4+3x^2-12x+8$ .

Answer to Problem 25PT

The rational zeroes of the function $f\left(x\right)=2x^4+3x^2-12x+8$ are

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$ .

Explanation of Solution

Given information:

The function is given as:

  $f\left(x\right)=2x^4+3x^2-12x+8$

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,

  $f\left(x\right)=2x^4+3x^2-12x+8$

Apply the rational zero theorem,

  $f\left(x\right)=2x^4+3x^2-12x+8=0$

If $\frac{p}{q}$ is a rational zero, then $p$ is a factor of the constant term $8$ and $q$ is a factor of the leading coefficient $2$ .

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

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

  $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$

Thus, the rational zeroes of the function $f\left(x\right)=2x^4+3x^2-12x+8$ are $\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{2}$