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