Answer
$\pm1,\pm4$.
Work Step by Step
Step 1. Treat $x^2$ as a unit and factor the equation to get $(x^2-1)(x^2-16)=0$
Step 2. For $x^2-1=0$ we have $x=\pm1$
Step 3. For $x^2-16=0$ we have $x=\pm4$
Step 4. The solutions are $x=\pm1$ and $x=\pm4$.
