## Precalculus (6th Edition) Blitzer

$(-2, -1) ∪(2, ∞)$
Consider the Polynomial Inequality as follows: $\frac{1}{4(x+2)}$$\gt$$-\frac{3}{4(x-2)}$ $\frac{1}{4(x+2)}$+$\frac{3}{4(x-2)}$$\gt$$0$ $\frac{4(x-2)+12(x+2)}{16(x+2)(x-2)}$$\gt$$0$ $\frac{4x-8+12x+24}{16(x+2)(x-2)}$$\gt$$0$ $\frac{16x+16}{16(x^{2}-4)}$$\gt$$0$ $\frac{16(x+1)}{16(x^{2}-4)}$$\gt$$0$ $\frac{(x+1)}{(x^{2}-4)}$$\gt$$0$ The graph of$f(x)$ is greater than 0 from $-2$ to $-1$ and from $2$ to $+∞$. Conclusion: Thus, the interval notation of the inequality is$(-2, -1) ∪(2, ∞)$ .