Answer
a. $ \pm 1, \pm 2,\pm 4$
b. sample: $ -1$ is a zero...(see "step by step")
c. $-2,-1,2$
Work Step by Step
see The Rational Zero Theorem:
... If $\displaystyle \frac{p}{q}$ is a zero of the polynomial $f(x) $with integer coefficients,
then $p$ is a factor of the constant term, $a_{0}$, and
$q$ is a factor of the leading coefficient, $a_{n}$.
------------------------
$f(x)=x^{3}+x^{2}-4x-4$
a. candidates for zeros, $\displaystyle \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2,\pm 4$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 4$
b. Try for $-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & 1 & -4 & -4\\
& & -1 & 0 & -4\\
& -- & -- & -- & --\\
& 1 & 0 & -4 & 0
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^{2} -4)$
c. Recognize the factor $(x^{2} -4)$ as a difference of squares:
$f(x)=(x+1)(x-2)(x+2).$
So, the zeros of f satisfy $f(x)=0$
$(x+1)(x-2)(x+2)=0$
$x\in\{-2,-1,2\}$
