Answer
See explanations.
Work Step by Step
Step 1. Given $f(x)=x^4-4x^3-20x^2+32x+12$, for $x=-1$, we have $f(-1)=(-1)^4-4(-1)^3-20(-1)^2+32(-1)+12=-35\lt0$
Step 2. For $x=0$, we have $f(0)=(0)^4-4(0)^3-20(0)^2+32(0)+12=12\gt0$
Step 3. As $f(-1)$ and $f(0)$ have opposite signs, based on the intermediate value theorem, there is a real zero between -1 and 0.