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        riddles >> easy >> Long polynomial
        
(Message started by: NickH on Mar 15^th, 2003, 2:58am)

Title: Long polynomial
Post by NickH on Mar 15^th, 2003, 2:58am

Find all real roots of x⁸ - 4x⁷ - 4x⁶ + 4x⁵ + 38x⁴ - 4x³ - 4x² + 4x + 1 = 0.

Title: Re: Long polynomial
Post by KicksGenius on Mar 15^th, 2003, 6:17am

There are 4 real roots to that 8th degree polynomial, I will not publish them at this time, because i only have the decimal approximations, and I assume that you would likeexact answers.

Title: Re: Long polynomial
Post by NickH on Mar 15^th, 2003, 6:33am

Yes, exact answers only, please! Having said that, if you have fairly accurate answers, you can probably guess the exact roots, and thereby factorize the polynomial. (But I'm not saying that's the best approach...)

Title: Re: Long polynomial
Post by KicksGenius on Mar 15^th, 2003, 4:26pm

[hide]Got them; 2-sqrt5, 2+sqrt5, 1-sqrt2, and 1+sqrt2[/hide]
I believe those are the exact real roots, if not, well, My name isn't Kicks.

Title: Re: Long polynomial
Post by SWF on Mar 17^th, 2003, 4:53pm

Maybe there is an easier way, but for me, finding it would take longer than factoring:

=(x⁴-6x³+6x²+6x+1)*(x⁴+2x³+ 2x²-2x+1)
=(x²-2x-1)*(x²-4x-1)*( x²+(1-sqrt3)x+2-sqrt3)(x²+(1+sqrt3)x+2+sqrt3)

The real roots are therefore 1+sqrt(2), 1-sqrt(2), 2-sqrt(5), 2+sqrt(5)

Title: Re: Long polynomial
Post by wowbagger on Mar 18^th, 2003, 6:08am

Well, I don't know how easy factoring that polynomial is for you, but it isn't a piece of cake for me. After all, I'm not a computer algebra system. ;)

On the other hand, I haven't really tried yet. Somehow I have the feeling that there's a trick to it...

Title: Re: Long polynomial
Post by NickH on Mar 18^th, 2003, 1:58pm

There is indeed a trick, of sorts. Here's a hint -- [hide]it involves making an appropriate substitution. (Substitutions are always "appropriate" in hindsight, aren't they?!)[/hide]

Title: Re: Long polynomial
Post by SWF on Mar 18^th, 2003, 5:42pm

I do not have access to computer algebra either, and waited to try this problem because I thought somebody would solve it that way. For me, factoring wasn't simple but was easier than figuring out the clever trick.