Inequality

We can accomplish this problem first by Multiplying by (z + 1)^2 on both sides. Then,we solve for x. 

$(3 - z)(z + 1) \geq 2(z + 4)(z + 1)^2\text{ and }z + 1 \neq 0\\ (z + 1)(2(z + 4)(z + 1) -(3-z)) \leq 0\text{ and }z + 1 \neq 0\\ (z + 1)(2z^2 +11z+5)\leq 0 \text{ and }z + 1 \neq 0\\ (z+1)(2z+1)(z+5)\leq 0 \text{ and }z + 1 \neq 0\\ z \leq -5 \text{ or }-1 < z \leq -\dfrac12$

Writing this in interval notation, we get

$z \in (-\infty, -5] \cup \left(-1, -\dfrac12\right]$\

This might be wrong...but it's probably correct LOL :)

Thanks! :)