# How do you factor x^3 - x^2 +x +3?

May 13, 2016

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

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

#### Explanation:

Notice that if the signs of the coefficients on the terms of odd degree are inverted then the sum of the coefficients is zero.

That is: $- 1 - 1 - 1 + 3 = 0$

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

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

The discriminant of the remaining quadratic factor is negative, but we can factor it with Complex coefficients by completing the square:

${x}^{2} - 2 x + 3$

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

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

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

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

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