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)#.