Answer
$x\displaystyle \in\{-2i, 2i, 1\}$
Work Step by Step
See Descarte's rule of sign:
It asserts that the number of positive roots is at most the number of sign changes in the sequence of polynomial's coefficients (omitting the zero coefficients), and that the difference between these two numbers is always even.
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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$f(x)=x^{3}+3x^{2}-4$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4$
$q:\qquad \pm 1$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2, \pm4$
b. $f(x)=x^3+3x^2-4$, there's one sign change therefore $f(x)$ has one positive root
$f(-x)=-x^3+3x^2-4$, there's two sign change therefore $f(x)$ has $2$ or $0$ negative roots
c. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &1& 3 & -4\\
& & 1 & 4\\
& -- & -- & -- & --\\
& 1 & 4 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^{2}+4)$
d. The zeros of $x^2+4=0$ are
$x=-2i, x=2i$
The solution set is:
$x\displaystyle \in\{-2i, 2i, 1\}$
