Answer
a. $\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2 \pm \frac{1}{2}$
b. $-2$ is a zero
c. $x\displaystyle \in\left\{\frac{-1 \pm i}{2}, -2\right\}$
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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$f(x)=2x^{3}+6x^{2}+5x+2$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1,\pm2 \pm \frac{1}{2}$
b. Try for $x=-2:$
$\begin{array}{lllll} \underline{-2}| & 2 & 6 & 5 & 2\\
& & -4 & -4 & -2\\
& -- & -- & -- & --\\
& 2 & 2 & 1 & |\underline{0} \end{array}$
$-2$ is a zero,
$f(x)=(x+2)(2x^{2} +2x+1)$
c. using quadratic formula, $x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}$. we can solve for the quadratic $ax^2+bx+c$, In our case of, $2x^{2} +2x+1$, $a=2, b=2, c=1$ $x=\frac{-2 \pm \sqrt {2^2-4\times 2 \times 1}}{2\times 2}$, $x=\frac{-1 \pm i}{2}$
The zeros of f satisfy $f(x)=0$
$x\displaystyle \in\left\{\frac{-1 \pm i}{2}, -2\right\}$
