Find the zero of f (x) = x^5-12x^2+32x

The zeroes of f (x) = x^5-12x^2+32x can be found by factoring,

f (x) = x^5-12x^2+32x = (x-8) (x-4)

By the zero product theorem, (x-8) = 0 or (x-4) = 0 which means

x=8 or x=4.

So the zeroes of f (x) are S={4,8}