# How do you solve 3r^5 + 15r^3 -18r = 0?

May 17, 2015

All the factors are divisible by $3 r$, so...

$0 = 3 {r}^{5} + 15 {r}^{3} - 18 r = 3 r \left({r}^{4} + 5 {r}^{2} - 6 r\right)$

Note that this means $r = 0$ is a root.

Next notice that $5 = 2 + 3$ and $6 = 2 \times 3$.

So ${r}^{4} + 5 {r}^{2} - 6 r = \left({r}^{2} - 2\right) \left({r}^{2} - 3\right)$

${r}^{2} - 2 = 0$ for $r = \pm \sqrt{2}$

${r}^{2} - 3 = 0$ for $r = \pm \sqrt{3}$

So the five roots are $r = 0$, $r = \pm \sqrt{2}$ and $r = \pm \sqrt{3}$.