Answer
$(-\infty,-2] \cup [2,\infty)$
Work Step by Step
The equation according to the text: $x^4-16\geq0\\(x-2)(x+2)(x^2+4)\geq0$
The zeros: $x+2=0\\x=-2\\$ or $x-2=0\\x=2$.
Then, according to the table ($x^2+4$ is non-negative because $x^2$ is non-negative) (where I used the zeros ($x=-2,2$) to divide the number line into different parts), the solution set: $(-\infty,-2] \cup [2,\infty)$ because the function is non-negative in this interval.