## Precalculus (6th Edition) Blitzer

The possible rational zeroes for the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-4x-4$ are $\pm 1,\pm 2,\pm 4$.
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 $\pm 1,\pm 2,\pm 4$.