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

$y = {\left(x - 3\right)}^{3} \left(x + 1\right) = \left(x - 3\right) \left(x - 3\right) \left(x - 3\right) \left(x + 1\right)$
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 $3$.