Polynomial Inequalities - Finding the solution set?

[nukeman]

Polynomial Inequalities - Finding the solution set??

Homework Statement

Solve the Inequality

2x^3 >-8x^2

Homework Equations

The Attempt at a Solution

Ok I am able to solve this by first figuring out the zeroes, and then testing with regions,

So my answer is x=0 and x = -4 are the zeros of f

BUT... now it says "What is the solution set, and using interval notation.

Using interval notation the answer is (-4,0) or (0, infinity)

HOW do I figure out the solution set and interval notation?

Thanks!

 

[rock.freak667]

You could sketch the graph which would make it easier to see. Since it is a cubic, with two roots, there's only two ways it can really look.

 

[nukeman]

we have to do it without graphing it.

 

[jbunniii]

I guess I'm not sure what your question is. You answered the question correctly: the solution set is any x in (-4,0) or (0, infinity). If you're looking for alternate ways to express this, you could write either

(-4,0)\cup (0,\infty)
or
\{x : -4 < x < 0 \textrm{ or } x > 0\}

 

[nukeman]

Yes I know that is the answer jbunniii, but why is it the answer. I don't understand how this is the answer.

Thanks

 

[Mark44]

Your inequality is equivalent to x³ + 4x² > 0, or x²(x + 4) > 0

For this product to be positive, there are only two possibilities:
1) Both factors are positive.
2) Both factors are negative.

If you explore both of these possibilities, you'll get your solution set.

 

[gb7nash]

You know that x = 0 and x = -4 are the only two possibilities for 2x³ = -8x².

The only thing left to do is to test an arbitrary x value before -4, between -4 and 0, and after 0. Let's say we want to test x = 1. In this case, we get 2(1)³ > -8(1)². We must then conclude that for all x > 0, we have 2x³ > -8x². Otherwise, there would have to exist another value of x > 0 such that 2x³ = -8x². This is impossible though since the only x values that work are x = 0 and x = -4.

The other two intervals are very similar. Test an arbitrary value between -4 and 0, and before -4 and see if 2x³ < -8x² or 2x³ > -8x²

 

[rock.freak667]

we have to do it without graphing it.

Mark44 and others gave the alternative ways of solving it. But I didn't mean to go get a graph plotter and do it but just make a general sketch of the graph, or was that what you meant? (no sketching or plotting)