Finding the equation of a cubic polynomial $f(x)$, then solving $f(x) = 120$. 
Determine algebraically the value(s) of $x$ where the cubic function that has zeroes at $2$, $3$, and $-5$, and passes through the point $(4,36)$ has a value of $120$.

It is clear that I need to solve for $x$ when $y=120$, but the wording has left me confused and not sure how to approach it. 

You should read it in two parts:

  
*
  
*Find the unique cubic polynomial $y = f(x)$ that has zeroes at $2$, $3$, and $-5$ and passes through the point $(4,36)$.
  
*Find all values of $x$ such that $f(x) = 120$.

As a hint for the first part, note that the zeroes tell us that the function has the form:
$$
f(x) = a(x - 2)(x - 3)(x + 5)
$$
where $a$ is some scaling factor that we can solve for by using the given point.
