How do you factor 6r^3-9r^3-4r+6?

Jun 2, 2015

I think the question contains an error: $- 9 {r}^{3}$ should be $- 9 {r}^{2}$

On the assumption that I'm correct, the cubic polynomial factors easily by grouping:

$6 {r}^{3} - 9 {r}^{2} - 4 r + 6$

$= \left(6 {r}^{3} - 9 {r}^{2}\right) - \left(4 r - 6\right)$

$= 3 {r}^{2} \left(2 r - 3\right) - 2 \left(2 r - 3\right)$

$= \left(3 {r}^{2} - 2\right) \left(2 r - 3\right)$

$= \left(\sqrt{3} r - \sqrt{2}\right) \left(\sqrt{3} r + \sqrt{2}\right) \left(2 r - 3\right)$