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

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Answer 1 · 2015-05-17T22:18:03

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

$0 = 3r^5+15r^3-18r = 3r(r^4+5r^2-6r)$

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

Next notice that $5 = 2+3$ and $6 = 2xx3$ .

So $r^4+5r^2-6r=(r^2-2)(r^2-3)$

$r^2-2 = 0$ for $r=+-sqrt(2)$

$r^2-3 = 0$ for $r=+-sqrt(3)$

So the five roots are $r=0$ , $r=+-sqrt(2)$ and $r=+-sqrt(3)$ .

How do you solve 3r^5 + 15r^3 -18r = 03r^5 + 15r^3 -18r = 0?