Proof help

[Chasney913]

Let's see. My teacher wants to introduce us to formal proofs, and has given us some small things based on that.
Anyway, I'm not really here about the proof itself. She decided to give a question that's for a "challenge", but is still due, so I'm not sure how that works.
I (obviously) don't want anyone to do the question for me. A push in the right direction would be great, though. My friends and I have been looking at it and we can't get a good foothold.

Without further ado:

f(x) is a polynomial, and has the following properties:
i. f(1) is 1
ii. f(2x)/f(x+1) = 8 - 56/(x+7) where both sides are not equal to zero

Prove:
f(x) is cubic
f(-1) is -5/21

I don't think we need to actually find f(x), because that would make both portions really easy.

The only I did was to change the RHS of ii into (8x)/(x+7), but that doesn't get me anywhere, other than possibly f(2x) = 0 when x = 0, meaning f(0) = 0. That means the constant term of the cubic is 0, but that doesn't seem to help.

Hopefully someone can help me out.

[DarkerLine]

It's a pretty cool problem!

One thing you know is that all polynomials can be factored (at least, over the complex numbers). So you can write f(x)=a(x-x₁)(x-x₂)(...)(x-x_n) - actually, it stops at n=3, being cubic, but you've yet to prove that. See what you can do from there.

You also will end up finding f(x) to answer the second part of the question.

[Chasney913]

Right, so I considered this, and I'll see if you agree. I said that there are three factors if cubic, and that one factor should be (x) if f(0) is 0. This leaves a quadratic, which I know how to factor. I also know those factors (conjugates) must multiply to give 1. Using that, I can show that the "a" coefficient and the "c" coefficient are equal. That leaves me with f(x) = ax^3 + bx^2 + ax.
Now, I need to solve for a and c. I'm thinking a system of equations, but I'm not sure what data to use. Maybe f(0) = 0 and f(1) = 1? I'll try that now.

EDIT: except for the fact that f(0) = 0 is not really a help in this case. a and b could be anything, and it'll still work.

[DarkerLine]

The coefficients of x^3 and x are not equal, so I'm not sure what you're doing. Furthermore, I don't think I follow your logic at all.

"I said that there are three factors if cubic" - you're trying to prove that f(x) is cubic. Assuming that it's cubic isn't going to do you any favors.

Given the factored form of f(x), with the x_i unknown, what does that tell you about f(2x)/f(x+1)?

[Chasney913]

The only thing I can see is that because the result has only x terms, and no higher power terms, some terms must cancel out.

I see what I was doing wrong in the a=c thing, so no need for me to explain that.

Maybe I should wait to post - I see we have a number of people reading. Maybe they have something useful as well.

[DarkerLine]

Final hint: where do you think the 8 in "8x/(x+7)" came from?

[Chasney913]

Well, it would need to be some multiplication of the top term 2x; that's the only way to get something on the top. At first it seems like it would be (2x) cubed, but that produces 8x^3, which is wrong. Although, it's possible it's the "a" term, or a combination between the factors and the "a" term. The "a" term gets canceled out in division, though, so it wouldn't really make a difference what it was.

[thornahawk]

Let's see...

1.) It is given that f[1] == 1.
2.) Using the identity f(2x)/f(x+1) = 8 - 56/(x+7) and replacing x with 0 and 1/2, we find that f[0] == 0 and f[3/2] == 15/8
3. ) Since 8 - 56/(x+7) is singular at x == -7, f(2x)/f(x+1) should also be singular at x == -7, and thus f[-6] == 0.

I think you already know what to do from there. ;)

thornahawk