# Thread: Solving third degree inequalities

1. ## Solving third degree inequalities

X^3 + 3x^2 - 16x < or = 48

I'm having trouble finding the way to solve this in my book. I don't remember how to factor an equation with ^3 please help!

2. So you have, equivalently,

$\displaystyle x^{3}+3x^{2}-16x-48\le 0.$

To find the roots of the LHS, use the rational roots theorem.

3. My mind is being blown. It's pretty hard to understand all these Wikipedia links. Trying to understand what you just linked but it's not happening. Trying to solve and then write the solution set in interval notation. Anyone have a way to describe this to a less intelligent person?

4. In the context of your problem, look for roots that look like divisors of 48 divided by divisors of 1 (both positive and negative). So, try

$\displaystyle \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 16, \pm 24, \pm 48.$

Any hits?

5. I would factor it: $\displaystyle x^2(x+3)-16(x+3)=(x^2-16)(x+3)$

6. Ah ok I'm starting to understand now..... but not really. p/q = You take the constant term and divide by the number in front of the highest exponent and then..... *explosion* mind blown. Then you plug in p/q for x and solve? Still not sure where all these numbers are coming from. At one point in the equation I need to get it in some sort of form like ( )( )( ) = 0. From this is how I get the interval notation. This is probably extremely easy too which is bumming me out.

7. O man I'm a total idiot. Factor by grouping! Thanks a bunch both of you. I'm going to go cry.

8. Originally Posted by street1030
O man I'm a total idiot. Factor by grouping! Thanks a bunch both of you. I'm going to go cry.
I wouldn't call yourself an idiot. I find factoring to be a very tricky task, most of the time. If I can have a more routine method, that's all to the good, in my opinion. Factoring, of course, is still necessary from time to time.

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# how to solve an inequality equatin in third degree

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