Answer
$x \in \{-2, 2\}$ with a multiplicity of $1$ and $x \in \{1\}$ with a multiplicty of $3$
Work Step by Step
See The Rational Zero Theorem:
If $\displaystyle \frac{p}{q}$ 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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$f(x)=x^5-3x^4-x^{3}+11x^{2}-12x+4$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm4,$
$q:\qquad \pm 1, $
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2, \pm4,$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &1& -3 & -1 & 11 & -12& 4\\
& & 1 &-2 & -3 & 8& -4\\
& -- & -- & -- & --\\
& 1&-2 & -3 & 8& -4 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(x^4-2x^3-3x^{2} +8x-4)$
Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &1& -2& -3 &8& -4\\
& & 1&-1 & -4 & 4\\
& -- & -- & -- & --\\
& 1 & -1 &-4 & 4 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)^2(x^3-x^2-4x+4)$
c. Solving for the quadrinomial, factorizing the quadrinomial:
$x^3-x^2-4x+4=x^2(x-1)-4(x-1)=(x^2-4)(x-1)$.
thus, $(x^2-4)(x-1)^3=0$.
$x^2-4=0, x^2=4, x=\pm2$ or $x-1=0, x=1$
therefore, the zeros are:
$x \in \{-2, 2\}$ with a multiplicity of $1$ and $x \in \{1\}$ with a multiplicty of $3$
