Question

How do you find the zeroes for `y=(x-3)^3(x+1)`?

Answer 1

`y = (x-3)^3(x+1) = (x-3)(x-3)(x-3)(x+1)`

This will be zero when any of the linear factors are zero, that is when `x=-1` or `x=3`.

If all of the linear factors are non-zero, then `y` will not be zero either.

This is a quartic - a polynomial of order `4` - in `x`. All quartic equations in one variable have a total of `4` roots, but some may be complex numbers or repeated. In your case, there is one root `x=-1` of multiplicity `1` and one repeated root `x=3` of multiplicity `3`.