Answer
$x\displaystyle \in\left\{ - i,i, -\frac{1}{2}\right\}$
The zero $-\frac{1}{2}$ has multiplicity $2$.
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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$f(x)=4x^4+4x^3+5x^2+4x+1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1$
$q:\qquad \pm 1, \pm 2,\pm 4$
$\displaystyle \frac{p}{q}:\qquad\pm 1 \pm \frac{1}{2}, \pm \frac{1}{4}$
Try for $x=-\frac{1}{2}$.
$\begin{array}{lllll}
\underline{-\frac{1}{2}}| & 4 & 4 & 5&4& 1\\
& & -2 & -1&-2& -1\\
& -- & -- & -- & --\\
& 4 &2& 4& 2 & |\underline{0}
\end{array}$
$-\frac{1}{2}$ is a zero,
$f(x)=(x+\frac{1}{2})(4x^3+2x^2+4x+2)$,
c. Factor the quadrinomial:
$4x^3+2x^2+4x+2=2x^2(2x+1)+2(2x+1)=(2x^2+2)(2x+1)=2(x^2+1)(2x+1)$.
Solve the equation $(2x+1)^2(x^2+1)=0$
$x\displaystyle \in\left\{ - i,i, -\frac{1}{2}\right\}$
The zero $-\frac{1}{2}$ has multiplicity $2$.
