Answer
$x \in \{-1-\sqrt {6}, -1+\sqrt {6}, -1, -4\}$ with a multiplicity of $1$
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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$f(x)=x^4+7x^{3}+9x^{2}-17x-20$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4, \pm5, \pm10, \pm20$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2, \pm4, \pm5, \pm10, \pm20$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| &1& 7 & 9 & -17 & -20\\
& & -1 &-6 & -3 & 20\\
& -- & -- & -- & --\\
& 1&6 & 3 & -20 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(x^3+6x^{2} +3x-20)$
Try for $x=-4:$
$\begin{array}{lllll}
\underline{-4}| &1& 6 & 3 & -20\\
& & -4 & -8 & 20\\
& -- & -- & -- & --\\
& 1 & 2 & -5 & |\underline{0}
\end{array}$
$-4$ is a zero,
$f(x)=(x+1)(x+4)(x^2+2x-5)$
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, $x^2+2x-5$, $x=\frac{-2\pm\sqrt {2^2-4 \times 1 \times(-5)}}{2\times 1}=\frac{-2 \pm 2\sqrt {6}}{2}=-1\pm \sqrt {6}$.
Thus, the zeros are :
$x \in \{-1-\sqrt {6}, -1+\sqrt {6}, -1, -4\}$ with a multiplicity of $1$
