Question

z^5-28z^3+27z=0 How do you use variable substitution and factoring to find all of the roots of this equation? I know how to do it if the z would have only been to the fourth power, but how do you do it with an odd numbered power like this? Thanks!!

Answer 1

First of all, one root is 0 and you can factor the equation to get
z(z^4 -28z^2 + 27) = 0
Looking at it, you can see that +1 and -1 are also roots. The term in parentheses is
(z^2-1)(z^2-27)
So that reduces it to
z(z-1)(z+1)(z^2-27)
The last term factors into
(z + 3sqrt3)(z - 3sqrt3)