Answer
$x=-\frac{1}{3}$ with a multiplicty of $2$ and $x=\frac{1}{2}$ with a multiplicty 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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$f(x)=18x^{3}+3x^{2}-4x-1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1,$
$q:\qquad \pm 1, \pm 2, \pm3, \pm6, \pm9, \pm18$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2},\pm\frac{1}{3}, \pm\frac{1}{6}, \pm\frac{1}{9}, \pm\frac{1}{18}$
b. Try for $x=\frac{1}{2}:$
$\begin{array}{lllll}
\underline{\frac{1}{2}}| & 18 & 3 & -4 & -1\\
& & 9 & 6 & 1\\
& -- & -- & -- & --\\
& 18 & 12 & 2 & |\underline{0}
\end{array}$
$\frac{1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(18x^{2} +2x+2)$
c. Factorize the trinomial factor $(18x^{2} +12x+2)$
(find two factors of $2(18)=36$ whose sum is $12):$
$(+6$ and $+6$)
$18x^{2} +12x+2=18x^{2} +6x+6x+2 \quad$...factor in pairs ...
$=6x(3x+1)+2(3x+1)=(3x+1)(6x+2)=(3x+1)(3x+1)2=2(3x+1)^2$
$f(x)=(2x-1)(3x+1)^2$
The zeros of f satisfy $f(x)=0$
$(2x-1)(3x+1)^2=0$
The zeros are:
$x=-\frac{1}{3}$ with a multiplicity of $2$ and $x=\frac{1}{2}$ with a multiplicity of $1$
