Answer
a. $ \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3, \pm 6$
b. $-1$ is a zero of the function
c. $x\in\left\{\frac{-3 \pm \sqrt {33}}{2}, -1\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ 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, $a_{n}$.
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$f(x)=x^{3}+4x^{2}-3x-6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm 3, \pm 6$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3, \pm 6$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| & 1 & 4 & -3 & -6\\
& & -1 & -3 & 6\\
& -- & -- & -- & --\\
& 1 & 3 & -6 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^{2} +3x-6)$
c. Using the quadratic formula, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$.
we can solve for the quadratic $ax^2+bx+c$,
In our case of, $x^2+3x-6$, $a=1, b=3, c=-6$
$x=\frac{-3 \pm \sqrt {3^2-4\times 1 \times (-6)}}{2\times 1}$,
$x=\frac{-3 \pm \sqrt {33}}{2}$
The zeros of f satisfy $f(x)=0$
$(x+\frac{3 + \sqrt {33}}{2})(x+\frac{3 - \sqrt {33}}{2})(x+1)=0$
$x\displaystyle \in\left\{\frac{-3 \pm \sqrt {33}}{2}, -1\right\}$
