Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=x^4+x^3-6x^2-20x-16$, for $x=3.2$, we have $f(3.2)=(3.2)^4+(3.2)^3-6(3.2)^2-20(3.2)-16=-3.8144\lt0$
Step 2. For $x=3.3$, we have $f(3.3)=(3.3)^4+(3.3)^3-6(3.3)^2-20(3.3)-16=7.1891\gt0$
Step 3. As $f(3.2)$ and $f(3.3)$ have opposite signs, based on the intermediate value theorem, there is a real zero between 3.2 and 3.3.