A concise solution to "find the range of $x$ in $(x-a)(x-b)(x-c)>0$ where $a<b<c$".

Question

Find the range of $x$ in $(x-a)(x-b)(x-c)>0$ where $a<b<c$.

Graphically, this is obvious; it's the "bump" between $a$ and $b$, and $c$ to $\infty$ (and beyond!).

Numerically, it's a bit tedious but gets the job done; I test all combinations of $x$ that are larger/smaller than $a$/$b$/$c$.

Algebraically, it's about as tedious as numerically; I rigorously solve the inequality and end up with a ton of little inequalities, then I throw out all the invalid and redundant inequalities and end up with $a<x<b$ and $c<x$.

Is there a short and sweet algebraic solution to this inequality?

Answer 1

$$ \begin{array}{c|lcr} x & (-\infty,a) & (a,b) & (b,c) & (c,\infty) \\ \hline \operatorname{sgn}(x-a) & -1 & +1 & +1 & +1 \\ \operatorname{sgn}(x-b) & -1 & -1 & +1 & +1 \\ \operatorname{sgn}(x-c) & -1 & -1 & -1 & +1 \\ \hline \operatorname{sgn}(x-a)(x-b)(x-c) & -1 & +1& -1 & +1 \\ \end{array} $$

Answer 2

Any $x$ satisfying your inequality will be such that $x\neq a,b,c$, so the LHS of the inequality will be a product of three nonzero reals. Since their product is positive, this means the number of negative reals (among those three) must be even.

For $0$ negative reals, you're in $x>c$.

For $2$ negative reals, you're in $a<x<b$.