- Basic inequalities like $x+10<0$ can be solved just by common sense.
- But longer polynomial inequalities, especially those involving multiplication, require us to consider the inequality in “cases”.

Case 1: both brackets evaluate to positive.

$x−5>0x>5x−3>0x>3⟹x∈(5,∞) $Case 2: both brackets evaluate to negative.

$x−5<0x<5x−3<0x<3⟹x∈(−∞,3) $Considering both cases:

$x∈(−∞,3)∪(5,∞)$- It’s tedious with just two cases. So instead of solving each case separately, we can use this thing called
**wavy curve method**and a couple of shortcuts to avoid solving each case.

## Steps:

- Factorize the inequality.
- Identify the critical points–points around which the sign of the expression changes.
- Plotting these critical points on a number line will divide the line into regions.

Now, normally we would consider each of these “regions” as a case and check if the numbers in the interval satisfy our inequality. But, instead of solving these cases, we can convert the expression into a familiar format such that we can find the sign of the interval without actually solving the expression.

~~Find the sign of our inequality in each of these regions.~~- Ensure that, in each of the factors, $(x−n),(c−x)$, our $x$ is always positive.

- Now starting from the right-most interval, we can start annotating our intervals with their signs.
- The right-most interval is always positive.
- When we cross over into the next interval, our sign may flip.
- Sign flips if the critical point’s bracket’s power is odd.
- Eg: $(x−5)_{2}(x+1)_{3}>0$. Here critical point 5 has an even power, so when we cross it, the sign will not flip. But while crossing -1, our sign will flip.

Here $(x_{2}+1)$ has been multiplied with 0 (hence ignored) as the term will always be positive.

$x∈(−∞,−3]∪(2,∞)∪{1}$