# How do you find all the zeros of h(x) = x³ - 3x² + 4x - 2?

Jun 5, 2016

$x = 1$ or $x = 1 \pm i$

#### Explanation:

$h \left(x\right) = {x}^{3} - 3 {x}^{2} + 4 x - 2$

Note that the sum of the coefficients is $0$. That is:

$1 - 3 + 4 - 2 = 0$

So $h \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ is a factor:

${x}^{3} - 3 {x}^{2} + 4 x - 2 = \left(x - 1\right) \left({x}^{2} - 2 x + 2\right)$

The remaining quadratic has negative discriminant, so only Complex zeros, but we can complete the square and use the difference of squares identity to find them:

${x}^{2} - 2 x + 2$

$= {x}^{2} - 2 x + 1 + 1$

$= {\left(x - 1\right)}^{2} - {i}^{2}$

$= \left(\left(x - 1\right) - i\right) \left(\left(x - 1\right) + i\right)$

$= \left(x - 1 - i\right) \left(x - 1 + i\right)$

So the other two zeros are $x = 1 \pm i$