Answer
$x \in \{-1-2i, -1+2i\}$ with a multiplicity of $1$ and $x \in \{-2\}$ with a multiplicity of $2$
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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$f(x)=x^4+6x^{3}+17x^{2}+28x+20$
a. candidates for zeros, $ \frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4, \pm5, \pm10, \pm20$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm4, \pm5, \pm10, \pm20$
b. Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| &1& 6 & 17 & 28 & 20\\
& & -2 &-8 & -18 & -20\\
& -- & -- & -- & --\\
& 1&4 & 9 & 10 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+2)(x^3+4x^{2} +9x+10)$
Try for $x=-2:$
$\begin{array}{lllll}
\underline{-2}| &1& 4 & 9 & 10\\
& & -2 & -4 & -10\\
& -- & -- & -- & --\\
& 1 & 2 & 5 & |\underline{0}
\end{array}$
$-2$ is a zero,
$f(x)=(x+2)^2(x^2+2x+5)$
c. Solving for the trinomial using quadratic formula for the quadratic equation of $ax^2+bx+c$, $x=\frac{-b\pm\sqrt {b^2-4ac}}{2a}$.
In this case, $x^2+2x+5$, $x=\frac{-2\pm\sqrt {2^2-4 \times 1 \times5}}{2\times 1}=\frac{-2 \pm4i}{2}=-1\pm2i$.
Thus, the zeros are :
$x \in \{-1-2i, -1+2i\}$ with a multiplicity of $1$ and $x \in \{-2\}$ with a multiplicity of $2$
