Chapter 4, Problem 51RE

To determine

To calculate:

The list of all potential rational zeroes of the given function:

  $f\left(x\right)=12x^8-x^7+6x^4-x^3+18x^2+x-3$

Answer to Problem 51RE

The list of all potential rational zeroes of $f\left(x\right)=12x^8-x^7+6x^4-x^3+18x^2+x-3$ is $\pm 1,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{4},\pm \frac{1}{6},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4}$ .

Explanation of Solution

Given information:

  $f\left(x\right)=12x^8-x^7+6x^4-x^3+18x^2+x-3$

Formula Used:

The potential rational zeroes is given by the following expression:

  $x=\pm \frac{p}{q}$

Where,

Quotients of the factors of the last term: $p$

The factors of the leading coefficient: $q$

Calculation:

The function is given as:

  $f\left(x\right)=12x^8-x^7+6x^4-x^3+18x^2+x-3$

Get that:

  $\begin{array}{l}p=3\\ q=12\end{array}$

The potential rational zeroes of the given function is:

  $\begin{array}{l}x=\pm \frac{p}{q}\\ x=\pm \frac{3}{12}\end{array}$

All the potential rational zeroes are:

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{3},\pm \frac{1}{6},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm \frac{3}{6},\pm \frac{3}{12}$

Simplify and remove duplicates, these are potential rational zeroes:

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{4},\pm \frac{1}{6},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4}$

Hence, the list of potential rational zeroes is:

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{3},\pm \frac{1}{4},\pm \frac{1}{6},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4}$