Answer
a.$\frac{p}{q}:\qquad \pm 1, \pm 2,\pm 4$
b. $-1$ is a zero of the function
c. $x\in \{-1,4\}$
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, $a_{n}$.
------------------------
$f(x)=x^{3}-2x^{2}-7x-4$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 4$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 4$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & -2 & -7 & -4\\
& & -1 & 3 & 4\\
& -- & -- & -- & --\\
& 1 & -3 & -4 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^{2} -3x-4)$
c. Factorize the trinomial factor $(x^{2} -3x-4)$
(find two factors of $4(-1)=-4$ whose sum is $3):$
$(4$ and $-1$)
$x^{2} -3x-4=x^{2}+x-4x-4 \quad$...factor in pairs ...
$=x(x+1)-4(x+1)=(x+1)(x-4)$
$f(x)=(x+1)^2(x-4)$
The zeros of f satisfy $f(x)=0$
$(x+1)^2(x-4)=0$
$x\displaystyle \in \{-1,4\}$
