How do you factor  x^3-3x^2+6x-18?

Sep 3, 2016

${x}^{3} - 3 {x}^{2} + 6 x - 18 = \left({x}^{2} + 6\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + 6 x - 18} = \left(x - \sqrt{6} i\right) \left(x + \sqrt{6} i\right) \left(x - 3\right)$

Explanation:

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms, so this will factor by grouping:

${x}^{3} - 3 {x}^{2} + 6 x - 18 = \left({x}^{3} - 3 {x}^{2}\right) + \left(6 x - 18\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + 6 x - 18} = {x}^{2} \left(x - 3\right) + 6 \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + 6 x - 18} = \left({x}^{2} + 6\right) \left(x - 3\right)$

That's as far as you can go using Real coefficients since ${x}^{2} + 6 \ge 6 > 0$ for any Real value of $x$. If you allow Complex coefficients then this factors further as:

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + 6 x - 18} = \left({x}^{2} - {\left(\sqrt{6} i\right)}^{2}\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + 6 x - 18} = \left(x - \sqrt{6} i\right) \left(x + \sqrt{6} i\right) \left(x - 3\right)$