# How do you factor completely 6x^3 - 10x^2 + 9x - 15?

Nov 19, 2015

Factor by grouping to find:

$6 {x}^{3} - 10 {x}^{2} + 9 x - 15$

$= \left(2 {x}^{2} + 3\right) \left(3 x - 5\right)$

$= \left(\sqrt{2} x - \sqrt{3} i\right) \left(\sqrt{2} x + \sqrt{3} i\right) \left(3 x - 5\right)$

#### Explanation:

$6 {x}^{3} - 10 {x}^{2} + 9 x - 15$

$= \left(6 {x}^{3} - 10 {x}^{2}\right) + \left(9 x - 15\right)$

=(2x^2(3x-5)+3(3x-5)

$= \left(2 {x}^{2} + 3\right) \left(3 x - 5\right)$

$2 {x}^{2} + 3$ has no simpler factors with Real coefficients, since ${x}^{2} \ge 0$ for all $x \in \mathbb{R}$, hence $2 {x}^{2} + 3 \ge 3 > 0$ for all $x \in \mathbb{R}$.

If we allow Complex coefficients then $2 {x}^{2} + 3$ can be treated as a difference of squares(!), hence:

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