# How do you find all the zeros of  X^3-3x^2+4x-2?

May 2, 2016

This cubic polynomial has zeros: $x = 1$, $x = 1 - i$, $x = 1 + i$

#### Explanation:

First note that the sum of the coefficients is zero. That is:

$1 - 3 + 4 - 2 = 0$

Hence $x = 1$ is a zero and $\left(x - 1\right)$ a factor:

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

The remaining quadratic factor only has Complex zeros, which we can find by completing the square and using the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(x - 1\right)$ and $b = i$ as follows:

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

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

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

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

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

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

Hence zeros $x = 1 \pm i$