Answer
$x\in\left\{\frac{-3}{4}, -\sqrt {2}i, \sqrt {2}i, 1 \right\}$
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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$f(x)=4x^4-x^{3}+5x^{2}-2x-6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3, \pm 6$
$q:\qquad \pm 1, \pm 2, \pm 4$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2},\pm \frac{1}{4},\pm \frac{3}{4}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &4 & -1 & 5 & -2 & -6\\
& &4 & 3 & 8& 6\\
& -- & -- & -- & --\\
& 4 & 3 & 8 & 6 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(4x^3+3x^{2} +8x+6)$,
To solve for the Quadrinomial,
Try for $x=\frac{-3}{4}:$
$\begin{array}{lllll}
\underline{\frac{-3}{4}}| & 4 & 3 & 8 & 6\\
& & -3 & 0 & -6\\
& -- & -- & -- & --\\
& 4 & 0 & 8 & |\underline{0}
\end{array}$
$\frac{-3}{4}$ is a zero,
$f(x)=(x+\frac{3}{4})(x-1)(4x^{2} +8)$,
c.To solve for the quadratic,
set it equal to $0$,
$4x^2+8=0$,
$x= \pm \sqrt {2}i$
$f(x)=(x+\frac{3}{4})(x-1)(x -\sqrt {2}i)(x+\sqrt {2}i)$
The zeros of f satisfy $f(x)=0$,
$(x+\frac{3}{4})(x-1)(x -\sqrt {2}i)(x+\sqrt {2}i)=0$
$x\displaystyle \in\left\{\frac{-3}{4}, -\sqrt {2}i, \sqrt {2}i, 1 \right\}$
