Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$ 
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.

Hint: Complete the square in your last expression $2x^2 + 12x + 16$.
Solution: 
Completing the square gives 
$$2x^2 + 12x + 16 = 2(x^2 + 6x + 4) = 2(x^2 + 2\cdot 3x + 3^2 - 1) = 2(x+3)^2 - 2$$
You already showed that your inequality is equivalent to $2x^2 + 12x + 16 \geq 0$.
Using the result of completing the sqare, we continue: $$2x^2 + 12x + 16 \geq 0 \\ \iff (x+3)^2 - 1 \geq 0 \\ \iff (x+3)^2 \geq 1 \\ \iff \left| x+3\right| \geq 1 \\ \iff \left| x - (-3)\right| \geq 1 \\ \iff \text{the distance between }x\text{ and }-3\text{ is at least }1 \\\iff x \leq -4 \text{ or } x\geq -2.$$
By the fact that $x = -4$ is forbidden, we can also write
$$x < -4 \text{ or }x\geq -2.$$
Comment:
I like your start where you multiply by $(x + 4)^2$ (and not just by $x+4$). In this way, you avoid the case-by-case analysis!
