Answer
a.$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}$
b. $\frac{1}{2}$ is a zero
c. $x\displaystyle \in\left\{\frac{-1 \pm \sqrt 5}{2}, \frac{1}{2}\right\}$
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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$f(x)=2x^{3}+x^{2}-3x+1$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1$
$q:\qquad \pm 1, \pm 2$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm \frac{1}{2}$
b. Try for $x=\frac{1}{2}:$
$\begin{array}{lllll} \underline{\frac{1}{2}}| & 2 & 1 & -3 & 1\\ & & 1 & 1 & -1\\ & -- & -- & -- & --\\ & 2 & 2 & -2 & |\underline{0} \end{array}$
$\frac{1}{2}$ is a zero,
$f(x)=(x-\frac{1}{2})(2x^{2} +2x-2)$
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, $2x^{2} +2x-2$, $a=2, b=2, c=-2$ $x=\frac{-2 \pm \sqrt {2^2-4\times 2 \times (-2)}}{2\times 2}$, $x=\frac{-1 \pm \sqrt 5}{2}$.
The zeros of f satisfy $f(x)=0$
$x\displaystyle \in\left\{\frac{-1 \pm \sqrt 5}{2}, \frac{1}{2}\right\}$
