#### Answer

$(-∞, -2) ∪ [-1, 2)$

#### Work Step by Step

Consider the Polynomial Inequality as follows:
$\frac{1}{4(x+2)}$$≤$$-$$\frac{3}{4(x-2)}$
$\frac{1}{4(x+2)}+\frac{3}{4(x-2)}≤0$
$\frac{4(x-2)+12(x+2)}{16(x+2)(x-2)}$$\leq$$0$
$\frac{4x-8+12x+24}{16(x^{2}-4)}$$\leq$$0$
$\frac{16(x+1)}{16(x^{2}-4)}$$\leq$$0$
$\frac{x+1}{x^{2}-4}≤0$
The graph of $ f(x)$ is less than 0 from $-∞$ to $-2$ and from $-1$ to $2$ and goes to the infinities in $2$ and $-2$, they never touch that point.
At this point, these infinities are not to be considered in the domain.
Conclusion: Thus, the interval notation of the inequality is$(-∞, -2) ∪ [-1, 2)$