# How do you factor completely  P(x)= x^3-2x^2+x-2?

Mar 4, 2018

Factored over the real numbers: $\left(x - 2\right) \left({x}^{2} + 1\right)$

Factored over the complex numbers: $\left(x - 2\right) \left(x + i\right) \left(x - i\right)$

#### Explanation:

We can factor by grouping:

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

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

This is all we can factor over the real numbers, but if we include complex numbers, we can factor the remaining quadratic even further using the difference of squares rule:

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

This gives the following complex factoring:

$\left(x - 2\right) \left(x + i\right) \left(x - i\right)$