$f(x) = -3x^4 - 4x^3 + 4x^2 + 96x -64$ is nonnegative on some interval 
Prove that the function $f(x) = -3x^4 - 4x^3 + 4x^2 + 96x -64$ is nonnegative only on some subinterval of $[0,3]$.

I thought about taking the first derivative and finding its zeroes but that seems very computational. Is there an easier way to solve this?
 A: By Descartes' rule of signs, the number of positive zeros of $f(x)$ is either $2$ or $0$.  Since $f(1) > 0$ while $f(0) < 0$ and $f(3) < 0$, we conclude there are exactly two positive zeros, one in the interval $(0,1)$ and one in $(2,3)$; the function is positive between those two zeros, and negative on the rest of $[0,3]$.
A: Since we are limited to "pre-calculus" methods, we can use the guidance of the Rational Zeroes Theorem to find one real zero of $ \  -3x^4 \ - \ 4x^3 \ + \ 4x^2 \ +  \ 96x  \ - \ 64 $ as  $ \ x \ = \ \frac{2}{3} \ $ , so this quartic polynomial may be factored as $ \ -3 \ (x \ - \ \frac{2}{3} ) \ (x^3 \ + \ 2x^2 \ - \ 32 ) \ $ , using polynomial or synthetic division.  
We have located a real zero with multiplicity 1 , so the curve for this function crosses the $ \ x-  $ axis there.  Some calculation with values of $ \ x \ $ in the vicinity will establish that the function is "passing from negative to positive" at this intercept. From what we know about the behavior of even-degree polynomials, and the fact that the leading coefficient is negative, the curve must have at least one turning point and a second $ \ x- $ intercept , since the function is negative for "large values of $ \ \vert \ x \ \vert \ $ ".  [I am adopting the language of textbooks such as Sullivan's here, which avoid saying things like "limits at infinity".]  (Further calculation will show that this intercept lies between $ \ x \ = \ 2 \ $ and $ \ x \ = \ 3 \ $ . )  We wish to show that there are no more $ \ x-$ intercepts. 
The curve for the function $ \ x^3 \ + \ 2x^2 \ = \ x^2 \ (x + 2) \ $ has a zero of multiplicity 1 at $ \ x \ = \  -2 \ $ and  a zero of multiplicity 2 at $ \ x \ = \ 0 \ $ .  So this curve crosses 
"from negative to positive" at $ \ x \ = \ -2 \ $ and remains non-negative for $ \ x \ > \ -2 \ $ .  We can easily show that for $ \ -2 \ < \ x  \ < \ 0 \ $ that $ \ 0 \ < \ 2x^2 \ < \ 8 \ $ , so certainly $ \ x^3 \ + \ 2x^2 \ < \ 8 \ $ in this interval. This cubic polynomial is positive for $ \ x \ > \ 0 \ $ , and all of its zeroes are accounted for.  If we now "shift downward" the curve to produce the curve for $ \ x^3 \ + \ 2x^2 \ - \ 32 \ $ , the positive region $ \ -2 \ < \ x  \ < \ 0 \ $ discussed above is now well below the $ \ x-$ axis , but the curve must still cross the $ \ x- $ axis somewhere for $ \ x \ > \ 0 \ $ .  This then is the remaining (aforementioned) $ \ x-$ intercept for $ \ -3 \ (x \ - \ \frac{2}{3} ) \ (x^3 \ + \ 2x^2 \ - \ 32 ) \ $ .  (The other two zeroes must then be complex numbers.)  This avoids the ambiguity of using the "Law of Signs" alone and the computation of the zeroes of the cubic factor by the "Cardano" formula.
