Answer
$x=1$ and $x=-2$
Work Step by Step
$x^{3} - 3x + 2 = 0$
Using the Rational Zero Theorem, we find the possible rational zeros as the ratio between the constant term and the leading coefficient:
$$PossibleZeros: \frac{+}{}\frac{1}{1}, \frac{+}{}\frac{2}{1}$$
and we use long division using the first possible zero (+1, also expressed as $x-1$):
...............$x^{2} + x - 2$
..............._______________
$(x - 1)|x^{3} + 0x^{2} - 3x + 2$
..............$-x^{3} + x^{2}$
...............________
............................$x^{2} - 3x$
............................-$x^{2} + x$
..........................._________
...................................$-2x + 2$
.......................................$2x - 2$
....................................._________
...............................................0
which means we can now re-write the original equation as:
$$(x-1)(x^{2} + x -2) = 0$$
Since two factors of $2$ that, when substracted, give $1$ are $2$ and $1$, we can also re-write it as:
$$(x-1)[(x+2)(x-1)]=0$$
$$(x-1)^{2}(x+2) = 0$$
Where $x-1=0$ and $x+2=0$, therefore $x=1$ and $x=-2$.
