How to solve inequality of the form $(x-a)(x-b)(x-c)\ge 0$? How to solve an inequality of the form $$f(x)=(x-a)(x-b)(x-c)\ge 0$$ for $$ a,b,c,x,f(x) \in \mathbb R $$ WITHOUT testing if an $f(x)$ within an interval between the roots is actually bigger or lesser than $0$?
 A: Let's assume that the roots are in increasing order. As the expression on the left is a cubic with positive leading coefficient, we know that the expression will be negative on the interval $(-\infty, a)$ and then change signs every time we come across a root of odd multiplicity. So, if the roots are all distinct - i.e. of multiplicity one - then the expression will be negative on $(-\infty, a)$, positive on $(a,b)$, negative on $(b,c)$ and finally positive again on $(c,\infty)$.
Note: Obviously we would include the roots when writing our solutions since we are allow equality in the original inequality statement.
A: Assume $a<b<c$ (if wlog $a=b$, then $(x-a)^2$ is only relevant to the inequality if $x=a$, otherwise it does nothing to the sign. see 2nd picture below).    
We use the following interval method (I don't know if it has a name) for solving inequalities of the form $$\frac{a(x-x_1)(x-x_2)\cdots(x-x_k)}{b(x-y_1)(x-y_2)\cdots(x-y_k)}\stackrel{\le }\ge 0$$ 
(or $\stackrel{<}>0$). For your inequality, we have $$(x-a)(x-b)(x-c)\ge 0\iff x\in[a,b]\cup[c,+\infty)$$   
I think the picture alone explains it well enough. The $x-x_i$ terms are linear and change sign at exactly one point, which is at $x_i$. The dots on $a,b,c$ in the picture should be deep black, and if inequality is $<$ or $>$ (i.e. not inclusive) or if the dots are when division by $0$ happens, we mark the dots by $\circ$ instead.

If the inequality was e.g. $(x-a)^2(x-b)(x-c)\le 0$ with $a<b<c$,   
the solution set would be $x\in\{a\}\cup [b,c]$ with diagram below:

I'm hoping this didn't sound vague.
