Approximate root-finding, by factorization or approximation?

Question

A previous question asked about a root of:
f(x) = 4x^7 +6x^6 +3x^5 +250x^4 +1000x^3 +75x^2 -20x +17

Can you estimate or generalize approximately about the roots, using only, pencil and paper, factorizations or approximations? (no calculators, computers, solvers or Newton-Raphson. See how far you can get using your head!).
Can you at least bound the regions in which each root lies, by identifying the zero-crossings of f?

Two ideas: a) since all terms other than "-20x" are positive, most roots will be negative.
Hence it might make it algebraically easier to substitute u→(-x) and consider
f(u) = -4u^7 +6u^6 -3u^5 +250u^4 -1000u^3 +75u^2 +20u +17

b) I'm thinking one promising way might be to regard it as the superposition of two (or more) polynomials, a high-order one and a lower-order one.

Your thoughts?

[ http://answers.yahoo.com/question/index?qid=20071215210154AAY3BhC ]

I think there can be at most one root 0 and the other six all negative.

Ok so it has three quadratic factors and at least one one real root. Can you at least just estimate the real root, or the quadratic factors?

I already said you don't need derivatives, you can use zero-crossings (method of false position) to get a bound on where the root(s) is(/are).
Clearly f(0) = +17 and f(-10) = very negative => f has at least one root in -10
And you can get a partial factorization approximation by
f(u) = -u^7 + (-3u^7 +6u^6 -3u^5) +250u^3 (u-4) +75u² +20u +17
= -u^7 - 3u^5 (u²-2u+1) +250u^3 (u-4) + 75u² +20u +17
= -u^7 - 3u^5 (u-1)² +250u^3 (u-4) + 75u² +20u +17
and so on. May be useful.

In fact f(0) = +17 and f(-1) <<0. So there's a bound on a root.
Now, does f have any other zero-crossings?

Answer 1

The only real root it has is x = - 0.324, because that's the only zero crossing it has, if you plot it out. That means all the rest of them are complex roots. It's actually possible to plot this function as a complex function, thereby roughly finding the other 6 complex roots, which are

(- 3.18 - 1.55 i)
(- 3.18 + 1.55 i)
(0.121 - 0.197 i)
(0.121 + 0.197 i)
(2.47 - 3.67 i)
(2.47 + 3.67 i)

I use software that lets you plot functions of a complex variable where the phase is denoted by color, so the roots are where points are the center of a rainbow fan of colors. In other words, sometimes doing it by brute numerical force is easier than trying to analyze a particular polynomial for factors. But I believe it's easier to understand properties of polynomials when they are studied as functions of complex variables.

As for "estimating roots without using calculators or software", it doesn't take long to realize that there has to be a zero point crossing between x = -1, and x = 0, and that there isn't any for x > 1, but trickier to prove that there aren't any as well for x < -1. The hardest thing to prove is that there aren't any other zero point crossings for 0 < x < 1. I have to admit I have no idea how to estimate complex roots, except to plot them out as I've described.

Answer 2

Honestly I wouldn't even know where to start with this...
How do you even know for sure it has more than one real root? Maybe the other six are all complex?

I can't even use the derivative since it's still a 6th degree... arg :)