Answer
$(-\infty,-2) \cup (-1,1) \cup (2,\infty)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$
where $f$ is a polynomial function.
$x^8-17x^4+16\gt 0,$ let's let $x^4=k$
$k^2-17k+16\gt 0,$
$k^2-k-16k+16\gt 0,$
$k(k-1)-16(k-1)\gt0,$
$(k-16)(k-1)\gt0,$ substituing back in $x^4=k$.
$(x^4-1)(x^4-16)\gt0,$
$(x^2+1)(x^2+4)(x-1)(x+1)(x-2)(x+2)\gt0,$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(x^2+1)(x^2+4)(x-1)(x+1)(x-2)(x+2) =0$
$x=-2$ or $x=-1$ or $x=1$ or $x=2$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \gt 0 ? \\
& &(a^2+1)(a^2+4)(a-1)(a+1)(a-2)(a+2) & \\
(-\infty, -2) & -5 & (+)(+)(-)(-)(-)(-)=(+) & T\\
(-2, -1) & -\frac{3}{2} & (+)(+)(-)(-)(-)(+)=(-) & F\\
(-1,1) & 0 & (+)(+)(-)(+)(-)(+)=(+) & T\\
(1, 2) & \frac{3}{2} & (+)(+)(+)(+)(-)(+)=(-) & F\\
(2,\infty) & 5 & (+)(+)(+)(+)(+)(+)=(+) & T
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $(-\infty,-2) \cup (-1,1) \cup (2,\infty)$
