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!

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- Feb 11th 2011, 09:56 AMstreet1030Solving 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! - Feb 11th 2011, 09:59 AMAckbeet
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. - Feb 11th 2011, 10:17 AMstreet1030
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?

- Feb 11th 2011, 10:19 AMAckbeet
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? - Feb 11th 2011, 10:24 AMPlato
I would factor it: $\displaystyle x^2(x+3)-16(x+3)=(x^2-16)(x+3)$

- Feb 11th 2011, 10:42 AMstreet1030
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.

- Feb 11th 2011, 10:44 AMstreet1030
O man I'm a total idiot. Factor by grouping! Thanks a bunch both of you. I'm going to go cry.

- Feb 11th 2011, 10:47 AMAckbeet