Chapter 2.4, Problem 36E

Solution

To find: The possible rational zeros of the polynomial $f\left(x\right)=6x^4-x^3-6x^2-x-12$ .

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

  $x=\frac{3}{2},-\frac{4}{3}$ are zeroes.

Given information:

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

Concept Used:

The Rational Zeros Theorem states: Factor the constant term and leading coefficient of the polynomial.

Calculation:

The constant term is $12$ .

The factors of $12:\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12$ .

The leading coefficient is $6$ .

The factors of $6:\pm 1,\pm 2,\pm 3,\pm 6$

Now find the possible rational zeroes:

  $\begin{array}{l}\frac{\pm 1}{\pm 1},\frac{\pm 2}{\pm 1},\frac{\pm 3}{\pm 1},\frac{\pm 4}{\pm 1},\frac{\pm 6}{\pm 1},\frac{\pm 12}{\pm 1},\\ \frac{\pm 1}{\pm 2},\frac{\pm 2}{\pm 2},\frac{\pm 3}{\pm 2},\frac{\pm 4}{\pm 2},\frac{\pm 6}{\pm 2},\frac{\pm 12}{\pm 2},\\ \frac{\pm 1}{\pm 3},\frac{\pm 2}{\pm 3},\frac{\pm 3}{\pm 3},\frac{\pm 4}{\pm 3},\frac{\pm 6}{\pm 3},\frac{\pm 12}{\pm 3},\\ \frac{\pm 1}{\pm 6},\frac{\pm 2}{\pm 6},\frac{\pm 3}{\pm 6},\frac{\pm 4}{\pm 6},\frac{\pm 6}{\pm 6},\frac{\pm 12}{\pm 6}\end{array}$

The zeros are: $\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12,\frac{\pm 1}{2},\frac{\pm 3}{2},\frac{\pm 1}{3},\frac{\pm 2}{3},\frac{\pm 4}{3},\frac{\pm 1}{6}$ .

There are total 24 possible zeros.

  $\begin{array}{c}f\left(\frac{3}{2}\right)=6{\left(\frac{3}{2}\right)}^4-{\left(\frac{3}{2}\right)}^3-6{\left(\frac{3}{2}\right)}^2-\left(\frac{3}{2}\right)-12\\ =0\\ f\left(-\frac{4}{3}\right)=6{\left(-\frac{4}{3}\right)}^4-{\left(-\frac{4}{3}\right)}^3-6{\left(-\frac{4}{3}\right)}^2-\left(-\frac{4}{3}\right)-12\\ =0\end{array}$

Hence, $x=\frac{3}{2},-\frac{4}{3}$ are zeroes.