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

Apr 7, 2016

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

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

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

#### Explanation:

All of the terms are divisible by $3 x {y}^{2}$, so separate that out as a factor first:

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

The remaining quadratic factor is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 1$ and $c = 1$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - \left(4 \cdot 1 \cdot 1\right) = 1 - 4 = - 3$

Since this is negative the quadratic has no factors with Real coefficients.

If we allow Complex coefficients then we can factor it further.

${x}^{2} - x + 1 = 0$ when

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{1 \pm \sqrt{- 3}}{2}$

$= \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

Hence we find:

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