Answer
$x\in \left\{-\frac{5}{2}, -\sqrt 3, \sqrt 3, 1 \right \}$
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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$f(x)=2x^4+3x^{3}-11x^{2}-9x+15$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 3, \pm5, \pm15$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 3,\pm 5, \pm 15, \pm \frac{1}{2}, \pm \frac{3}{2},\pm \frac{5}{2},\pm \frac{15}{2}$
b. Try for $x=1:$
$\begin{array}{lllll}
\underline{1}| &2 & 3 & -11 & -9 & 15\\
& &2 & 5 & -6& -15\\
& -- & -- & -- & --\\
& 2 & 5 & -6 & -15 & |\underline{0}
\end{array}$
$1$ is a zero,
$f(x)=(x-1)(2x^3+5x^{2} -6x-15)$,
To solve for the Quadrinomial,
Try for $x=\frac{-5}{2}:$
$\begin{array}{lllll}
\underline{\frac{-5}{2}}| & 2 & 5 & -6 & -15\\
& & -5 & 0 & 15\\
& -- & -- & -- & --\\
& 2 & 0 & -6 & |\underline{0}
\end{array}$
$\frac{-5}{2}$ is a zero,
$f(x)=(x+\frac{5}{2})(x-1)(2x^{2} -6)$,
c. To solve for the quadratic, set it equal to $0$,
$2x^2-6=0$,
$x= \pm \sqrt 3$
$f(x)=(x+\frac{5}{2})(x-1)(x -\sqrt 3)(x+\sqrt 3)$
The zeros of f satisfy $f(x)=0$,
$(x+\frac{5}{2})(x-1)(x -\sqrt 3)(x+\sqrt 3)=0$
$x\displaystyle \in\left\{\frac{-5}{2}, -\sqrt 3, \sqrt 3, 1 \right\}$
