What are all the real zeros of #y=(x-12)^3-10#?

1 Answer
Aug 20, 2017

Answer:

The real zero is:

#x = 12+root(3)(10)#

Explanation:

Given:

#y = (x-12)^3-10#

The real zero is:

#x = 12+root(3)(10)#

The other zeros are both non-real complex:

#x = 12+omega root(3)(10)#

#x = 12+omega^2 root(3)(10)#

where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.