Answer
Possible rational roots:
$ \displaystyle \pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2}$
Actual rational roots:
$ -3, -\displaystyle \frac{3}{2}, -1,$ and $2.$
(see image below)
Work Step by Step
By the Rational Zero Theorem,
if there is a rational root$, \displaystyle \frac{p}{q}$, of a polynomial equation,
then p is a factor of the constant term, $a_{o}$ and
q is a factor of the coefficient of the leading term $a_{n}$.
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$p:\quad \pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18$
$q:\quad \pm 1,\pm 2,$
$\displaystyle \frac{p}{q}:\quad \pm 1,\pm 2,\pm 3,\pm 6,\pm 9,\pm 18,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{9}{2}$
As a graphing utility, I used desmos.com online calculator.
Adjust the viewing window and step with the dialog box that opens when
we press the "wrench" button in the right upper corner (see image below)
On the resulting graph, we see that actual roots are
$ -3, -\displaystyle \frac{3}{2}, -1,$ and $2.$