real roots of f = 2(b^2)x^3 + abx^2 - [(a^2)+(3b^2)+1] - 3ab

[BrainMan]

I'm stuck on this problem:

Let a and b be fixed non-zero real numbers and let f(x) = 2(b^2)x^3 + abx^2 - [(a^2) + (3b^2) + 1] - 3ab. Prove that f(x) = 0 has three real roots and one of these roots must lie in the interval [-1,1].

I see that you can factorize it, but I'm not really sure what to do after that.
Thank you for your help.

 

[stapel]

If you can factor this, why not start there? (Is there supposed to be an "x" after the bracketed bit?)

Also, what theorems do you have to work with?

Eliz.

 

[BrainMan]

I'm pretty sure I know how to prove that one root lies in the interval [-1,1]. I find f(-1) and f(1) and then use the Intermediate Value Theorem since f(-1) is less than zero and f(1) is greater than zero. However, how do I show that there are three real roots? Can someone just show me this?

By the way, there should be an x after the brackets in my original function f(x).

I'm taking an elementary analysis class.

 

[Deleted member 4993]

Re: real roots of f = 2(b^2)x^3 + abx^2 - [(a^2)+(3b^2)+1] -

I'm stuck on this problem:

Let a and b be fixed non-zero real numbers and let f(x) = 2(b^2)x^3 + abx^2 - [(a^2) + (3b^2) + 1]x - 3ab. Prove that f(x) = 0 has three real roots and one of these roots must lie in the interval [-1,1].

I see that you can factorize it,

What are the factors?

but I'm not really sure what to do after that.
Thank you for your help.