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

Jan 10, 2018

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

$\textcolor{w h i t e}{{x}^{3} - 2 {x}^{2} + 3} = \left(x + 1\right) \left(x - \frac{3}{2} - \frac{\sqrt{3}}{2} i\right) \left(x - \frac{3}{2} + \frac{\sqrt{3}}{2} i\right)$

#### Explanation:

Given:

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

Note that:

$f \left(- 1\right) = - 1 - 2 + 3 = 0$

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

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

The remaining quadratic has negative discriminant, so can only be factored further using complex coefficients:

$4 \left({x}^{2} - 3 x + 3\right) = 4 {x}^{2} - 12 x + 12$

$\textcolor{w h i t e}{4 \left({x}^{2} - 3 x + 3\right)} = {\left(2 x\right)}^{2} - 2 \left(2 x\right) \left(3\right) + {3}^{2} + 3$

$\textcolor{w h i t e}{4 \left({x}^{2} - 3 x + 3\right)} = {\left(2 x - 3\right)}^{2} + {\left(\sqrt{3}\right)}^{2}$

$\textcolor{w h i t e}{4 \left({x}^{2} - 3 x + 3\right)} = {\left(2 x - 3\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$\textcolor{w h i t e}{4 \left({x}^{2} - 3 x + 3\right)} = \left(\left(2 x - 3\right) - \sqrt{3} i\right) \left(\left(2 x - 3\right) + \sqrt{3} i\right)$

$\textcolor{w h i t e}{4 \left({x}^{2} - 3 x + 3\right)} = \left(2 x - 3 - \sqrt{3} i\right) \left(2 x - 3 + \sqrt{3} i\right)$

So:

${x}^{2} - 3 x + 3 = \left(x - \frac{3}{2} - \frac{\sqrt{3}}{2} i\right) \left(x - \frac{3}{2} + \frac{\sqrt{3}}{2} i\right)$