Answer
The possible rational zeroes for the function \[f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15\] are \[\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}\].
Work Step by Step
Here, the constant term is $15$ and the leading coefficient is 2.
The factors of the constant term, $15$ are $\pm 1,\pm 3,\pm 5,\pm 15$ and the factors of the leading coefficient, 2 are $\pm 1,\pm 2$.
So, the list of all possible rational zeroes is calculated by the formula:
$\begin{align}
& \text{Possible rational zeroes}=\frac{\text{Factors of the constant term}}{\text{Factors of the leading coefficient}} \\
& =\frac{\text{Factors of 15}}{\text{Factors of 2}} \\
& =\frac{\pm 1,\pm 3,\pm 5,\pm 15}{\pm 1,\pm 2} \\
& =\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}
\end{align}$
Therefore, there are total sixteen possible rational zeroes for the function $f\left( x \right)=2{{x}^{4}}+3{{x}^{3}}-11{{x}^{2}}-9x+15$ that are $\pm 1,\pm 3,\pm 5,\pm 15,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$.
