Answer
$\left[-2,-\frac{3}{2}\right] \cup \left[\frac{3}{2}, 2\right]$
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.
$4x^4-25x^2+36 \leq 0$,
$4x^4-16x^2-9x^2+36 \leq 0$,
$4x^2(x^2-4)-9(x^2-4) \leq 0$,
$(4x^2-9)(x^2-4) \leq 0$,
$(2x-3)(2x+3)(x-2)(x+2) \leq 0$
$f(x)=(2x-3)(2x+3)(x-2)(x+2)$
2. Solve the equation $f(x)=0$. The real solutions are the boundary points.
$(2x-3)(2x+3)(x-2)(x+2)=0$
$x=\frac{3}{2}$ or $x=-\frac{3}{2}$ or $x=2$ 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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\
& &(2a-3)(2a+3)(a-2)(a+2)& \\
(-\infty, -2) & -4 & (-)(-)(-)(-) & F\\
(-2,-\frac{3}{2}) & -1.75 & (-)(-)(-)(+) & T\\
(-\frac{3}{2}, \frac{3}{2}) & 0 & (-)(+)(-)(+) & F\\
(\frac{3}{2}, 2) & 1.75 & (+)(+)(-)(+) & T\\
(2,\infty) & 4 & (+)(+)(+)(+) & F
\end{array}$
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: $\left[-2,-\frac{3}{2}\right] \cup \left[\frac{3}{2}, 2\right]$
