Answer
$x=\pm2$
Work Step by Step
The zeros of the function are the $x$-intercepts of this function. The $x$-intercepts can be found by equating the function expression to $0$ and solving for $x$:
$$f(x)=4x^2-16$$
$$4x^2-16=0$$
$$4x^2=16$$
$$x^2=4$$
$$x=\pm\sqrt{4}=\pm2$$