Answer
The possible rational zeroes for the function \[f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-{{x}^{2}}+19x+6\] are \[\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}\].
Work Step by Step
Here, the constant term is $6$ and the leading coefficient is 3.
The factors of the constant term, $6$ are $\pm 1,\pm 2,\pm 3,\pm 6$ 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 is:
$\begin{align}
& \text{Possible rational zeroes}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \\
& =\frac{\text{Factors of 6}}{\text{Factors of 3}} \\
& =\frac{\pm 1,\pm 2,\pm 3,\pm 6}{\pm 1,\pm 3} \\
& =\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}
\end{align}$
Therefore, there are total twelve possible rational zeroes for the function $f\left( x \right)=3{{x}^{4}}-11{{x}^{3}}-{{x}^{2}}+19x+6$ that are $\pm 1,\pm 2,\pm 3,\pm 6,\pm \frac{1}{3},\pm \frac{2}{3}$.
