Answer
The possible rational zeroes for the function \[f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8\] are \[\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3}\].
Work Step by Step
Here, the constant term is $8$ and the leading coefficient is 3.
The factors of the constant term, $8$ are $\pm 1,\pm 2,\pm 4,\pm 8$ and the factors of the leading coefficient, 3 are $\pm 1,\pm 3$.
So, the list of all possible rational zeroes is calculated by the formula:
$\begin{align}
& \text{Possible rational zeroes}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \\
& =\frac{\text{Factors of 8}}{\text{Factors of 3}} \\
& =\frac{\pm 1,\pm 2,\pm 4,\pm 8}{\pm 1,\pm 3} \\
& =\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3}
\end{align}$
Therefore, there are total sixteen possible rational zeroes for the function $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-3{{x}^{2}}-6x+8$ that are $\pm 1,\pm 2,\pm 4,\pm 8,\pm \frac{1}{3},\pm \frac{2}{3},\pm \frac{4}{3},\pm \frac{8}{3}$.
