Question

How do you find the zeros of `y=x^3-4x^2+8`?

Answer 1

One zero is `2` the other two are `1 pm sqrt(5)`

Explanation:

(2)^3 - 4(-2)^2 + 8 = 8 - 16 + 8 = 0
Then you factor out (x-2) to get
`(x-2)(x^2 -2x -4)` whose zeros are
`x = [2 +- sqrt(4 +4*4)]/2 = 1 pm sqrt(5)`