Question

How do you find the zeros of `y=2x^3-8x^2+18x-72`?

Answer 1

This can be factored as `y = 2x^2(x - 4) + 18(x - 4) = (2x^2 +18)(x - 4)`.

Setting as `y =0` and solving:

`0 = (2x^2 + 18)(x - 4)`

`x = +-3i and 4`

Hopefully this helps!