Answer
$\quad x\displaystyle \in\left\{-1-\sqrt{2},-1+\sqrt{2},\frac{2}{3}\right\}$
Work Step by Step
Factoring in pairs is not obvious here, so we try to find rational zeros of
$f(x)=3x^{3}+4x^{2}-7x+2$
Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 2}{\pm 1,\pm 3}$
Testing with synthetic division, $x\pm 1$ both fail (not zeros),
try $x-\displaystyle \frac{2}{3}$
$\left.\begin{array}{l}
2/3 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr}
3 &4 &-7 &2 \\\hline
&2 &4 & -2 \\\hline
3& 6 &-3 & |\ \ 0 \end{array}$
$f(x)=(x-\displaystyle \frac{2}{3})( 3x^{2}+6x-3)$
$=(x-\displaystyle \frac{2}{3})\cdot 3( x^{2}+2x-1)$
$=(3x-2)( x^{2}+2x+1-1-1)\qquad$... completing a square
$=(3x-2)[( x+1)^{2}-2]\qquad$ ... difference of squares in brackets
$=(3x-2)(x+1+\sqrt{2})(x+1-\sqrt{2})$
$ f(x)=0\quad$when$\quad x\displaystyle \in\left\{-1-\sqrt{2},-1+\sqrt{2},\frac{2}{3}\right\}$
