Answer
$x\in\left\{-\frac{1}{3}, -1, 2, 3\right\}$
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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$f(x)=3x^4-11x^{3}-x^{2}+19x+6$
a. Candidates for zeros, $\frac{p}{q}:$
$p:\qquad \pm 1, \pm 2, \pm3, \pm6$
$q:\qquad \pm 1, \pm 3$
$\displaystyle \frac{p}{q}:\qquad \pm 1, \pm 2,\pm 3, \pm 6, \pm \frac{1}{3}, \pm \frac{2}{3}$
b. Try for $x=-1:$
$\begin{array}{lllll}
\underline{-1}| &3 & -11 & -1 & 19 & 6\\
& &-3 & 14 & -13& -6\\
& -- & -- & -- & --\\
& 3 & -14 & 13 & 6 & |\underline{0}
\end{array}$
$-1$ is a zero,
$f(x)=(x+1)(3x^3-14x^{2} +13x+6)$,
To solve for the Quadrinomial,
Try for $x=2:$
$\begin{array}{lllll}
\underline{2}| & 3 & -14 & 13 & 6\\
& & 6 & -16 & -6\\
& -- & -- & -- & --\\
& 3 & -8 & -3 & |\underline{0}
\end{array}$
$2$ is a zero,
$f(x)=(x+1)(x-2)(3x^{2} -8x-3)$,
c. Factorize the trinomial factor $(3x^{2} -8x-3)$
$3x^{2} -8x-3=3x^{2} +x-9x-3 \quad$...factor in pairs ...
$=x(3x+1)-3(3x+1)=(3x+1)(x-3)$
$f(x)=(x-2)(3x+1)(x-3)(x+1)$
The zeros of f satisfy $f(x)=0$,
$(x-2)(3x+1)(x-3)(x+1)=0$
$x \in\left\{-\frac{1}{3}, -1, 2, 3\right\}$
