Answer
$x\displaystyle \in\left\{-\frac{1}{2}, \frac{-1 - i}{2},\frac{-1 + i}{2}, 1\right\}$
Work Step by Step
See The Rational Zero Theorem:
If $\frac{p}{q}$ is a zero of the polynomial $f$ 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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$f(x)=4x^4+2x^3-2x^2-3x-1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1$
$q:\qquad \pm 1, \pm 2,\pm 4$
$\displaystyle \frac{p}{q}:\qquad\pm 1 \pm \frac{1}{2}, \pm \frac{1}{4}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| & 4 & 2 & -2 & -3& -1\\
& & 4 & 6 & 4& 1\\
& -- & -- & -- & --\\
& 4& 6& 4& 1 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(4x^3+6x^{2}+4x +1)$
Try for $x=-\frac{1}{2}$.
$\begin{array}{lllll}
\underline{-\frac{1}{2}}| & 4 & 6 & 4& 1\\
& & -2 & -2& -1\\
& -- & -- & -- & --\\
& 4 & 4& 2 & |\underline{0}
\end{array}$
$-\frac{1}{2}$ is a zero,
$f(x)=(x-1)(x+\frac{1}{2})(4x^2+4x+2)$,
c. Solve for the trinomial using the quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$
in this case, $4x^2+4x+2$, $x=\frac{-4 \pm \sqrt {4^2-4 \times4 \times 2}}{2 \times 4}=\frac{-4\pm4i}{8}=\frac{-1\pm i}{2}$
$x\displaystyle \in\left\{-\frac{1}{2}, \frac{-1 - i}{2},\frac{-1 + i}{2}, 1\right\}$
