Solve $\frac{(x-1)^{204}(x+3)^5(x-4)^{2015}}{(x+5)^{102}}\ge 0$ Solve $\frac{(x-1)^{204}(x+3)^5(x-4)^{2015}}{(x+5)^{102}}\ge 0$ Just wanted to share a nice and quick technique i learnt for such problems.
 A: Immediately plot the critical points -5__-3____1__4 on a number line.Check for validity of each range by taking a number in that range.For x<-5 take -6;it holds true.For (-5,-3);-4 satisfies the inequality so x this range holds true.For (-3,1) take 0;the inequality does not hold.For (1,4) take 2 and check;it again does not hold.For x>4 take 5 to check;it holds.Now check the critical points ${-5,-3,1,4}$.Out of these ${-3,1,4}$ hold.So the solution of x is: $x<-5,-5<x<=-3,x>=4,x=1$
A: For the equality, $x=1,-3,4$ 
For $\dfrac{(x-1)^{204}(x+3)^5(x-4)^{2015}}{(x+5)^{102}}>0,$
As $x$ is real, the problem reduces to $(x+3)(x-4)>0\iff x>4$ or $x<-3$
But the denominator is undefined  at $x+5=0$
A: By the rule of signs, factors with  an even exponent are $\geq 0$ and those with an odd exponent have the same sign as with exponent $1$. Hence, if $x\neq -5\,$:
\begin{align*}
 \frac{(x-1)^{204}(x+3)^5(x-4)^{2015}}{(x+5)^{102}}\geq 0&\iff (x+3)(x-4)\geq 0\enspace\text{or}\enspace x=1\\
&\iff x\leq -3\enspace\text{or}\enspace x=1\enspace\text{or}\enspace x\geq 4
\end{align*}
Thus the set $\mathcal S$ of solutions is:
$$\mathcal S=(-\infty,-5)\cup(-5,-3]\cup\{1\}\cup[4,+\infty).$$
