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.

No. You should have $2(x+4)^2$ on the right after multiplying. Then $$0\le2(x+4)^2-4(x+4)\\0\le(x+4)^2-2(x+4)\\0\le(x+4)\bigl((x+4)-2\bigr)\\0\le(x+4)(x+2).$$ Now, that last inequality is true whenever $x+4$ and $x+2$ have the same sign or one of them is $0$--that is, whenever $x+4\le0$ ($x\le-4$) or $x+2\ge0$ ($x\ge -2$)--but we can't allow $x=-4$ in our original inequality. Hence, our given inequality is true whenever $x<-4$ or $x\ge -2$.
