Finding the range of a $y=-x^2(x+5)(x-3)$ without calculus? I was helping a precalculus student with this question. The graph wasn't given. My only idea was to find the inverse and try to find its domain. When trying to find the inverse, I arrived at $0=y^4+2y^3-15y^2+x$ and I was hoping to be able to factor it. However, besides using the quartic formula, I'm not sure how to do this.
 A: Well, the function is always negative if $x<-5$ or $x>3$, is equal to zero when $x$ is $-5$ or $3$, and grows in magnitude without bound as $x$ increases past $3$, so the range includes $(-∞,0]$.
Between $-5$ and $3$ the function is positive (except at $0$), so it's a question of finding the maximum achieved in this range. Personally, if I wasn't using calculus, I would graph the function with a graphing calculator, observe that the maximum is about $119.75$, and conclude that the range is approximately $(-\infty, 119.75]$.
A: First we have to find the zeros of the eqn $y=-(x^2)(x+5)(x-3)$. So from the 
given eqn we get the zeros which are 0 of multiplicity 2,-5 of 1 and 3 of 1.
1.$y>0$ when $-5<x<3$
2.$y<0$ when $-\infty<x<-5\;\;and\;\;3<x<\infty$
3.As $x$ tends to $-\infty$, $y$ tends to $-\infty$ and $x$ tends to +infinty, $y$ tends to -infinity.
Now we have to calculate the max value of $y$.
$y=-(x^2)(x+5)(x-3)$
$\implies$$y=-(x^2)({x^2}+2x-15)$
$\implies$$y=-({x^4}+2{x^3}-15{x^2})$
Then make this term ${x^4}+2{x^3}-15{x^2}$ into of the form
$y=-({f(x)^4})+K_1\;\;where\;\;K_1>0$
Or
$y=-({f(x)^4}+{g(x)^2})+K_2\text{ where }K_2>0$
Then you will get $K_1=K_2$, the max value of $y$.
Then you will get the $R(y)=(-\infty,K_1\;\;OR\;\;K_2$]
