# Thread: Polynomial inequalities

1. ## Polynomial inequalities

Solve the inequality:

$y^3 - 6y^2 + 8y < 0$

So here is what I did, just following the steps to follow a normal equation

y (y² - 6y + 8) < 0
y (y - 2) (y - 4) < 0

So what are the answers and how do i express them?

2. See here.

3. Thanks but I dont get all that stuff, with the infinity signs and stuff. Is there a easier method?

4. Hi
Originally Posted by Power
Solve the inequality:

y^3 - 6y² + 8y < 0

So here is what I did, just following the steps to follow a normal equation

y (y² - 6y + 8) < 0
y (y - 2) (y - 4) < 0

So what are the answers and how do i express them?
A product is negative iff it has an odd number of negative term(s). In this case, there are three terms. ( $y$, $y-2$ and $y-4$) As the only odd numbers in $\{1,2,3\}$ are 1 and 3, the product is negative if one term is negative or if the three are negative. Does it help ?

5. Originally Posted by flyingsquirrel
Hi

A product is negative iff it has an odd number of negative term(s). In this case, there are three terms. ( $y$, $y-2$ and $y-4$) As the only odd numbers in $\{1,2,3\}$ are 1 and 3, the product is negative if one term is negative or if the three are negative. Does it help ?
So whats the solution? I mean a solution expressed without a graph.

6. I suggest you to take another look to my link provided, it's the best way to tackle your problem, besides it's faster.