Question

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

Answer 1

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}