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Zeros of a polynomial
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arp
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#1
Zeros of a polynomial

How do you find the zeros of a polynomial of even degree above 4 whose graph lies above the x-axis?

PostPosted: Fri Feb 18, 2005 9:13 pm  Back to top 
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EFuzzy
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#2
There's no fool proof method for finding the (exact) complex zeroes, only some tricks for special cases.
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PostPosted: Sat Feb 19, 2005 5:47 am  Back to top 
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arp
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#3
hm could you give me an example of some of these tricks?

PostPosted: Sat Feb 19, 2005 7:33 am  Back to top 
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p_adic
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#4
Polynomials

Examples:

P(x) = x^6 + x^4 + x^2 + 1 is always positive, for real x to find complex zeros note that

P(x) = (x^8 -1)/(x^2 - 1) and hence P(x) is zero when x^8 - 1 = 0 and x^2 - 1 is nonzero.


Q(x) = x^8 + 2x^4 + 1 factors as (x^4 + 1)^2 and so the zeros are the same as those of x^4 + 1 = 0.


R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:

x^4 + 5(x + 1/x) + 1/x^4 = 0

Let u = x + 1/x

Then u^2 = x^2 + 1/x^2 + 2,

(u^2 - 2)^2 = x^4 + 1/x^4 + 2

So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is

(u^2 -2)^2 - 2 + 5u = 0

which can be solved for u by the quartic formula and then for u.

PostPosted: Sat Feb 19, 2005 6:17 pm  Back to top 
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goodyfresh741
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#5
Unsolvability of quintic

Don't go around looking for a solution (by algebraic methods) to the generalized quintic equation. In the 1800's (I'm not sure exactly when) Abel proved that it's impossible to find an algebraic solution to a general equation of degree >=5. In other words, choose any equation of degree >=5, and (with almost 100 % likelihood) it is impossible through a finite number of algebraic operations (addition and subtraction of polynomials from both sides, plus root-taking operations and such) to solve the equation. That can only be done in special cases. If you're interested in reading more about it, there is a "general solution" to the quintic equation using something called Jacobian theta functions (you can also use Weirstrassian theta functions for that purpose) but it's quite high-level math and won't do anything for you on a high school math contest.
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PostPosted: Mon Feb 21, 2005 12:49 pm  Back to top 
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jli
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#6
find all the factors of the coeficient of the one with the largest power

find the factors of the number at the end

divide them in all the possible ways and put a plus/minus in fron of the fractions and those are the possible roots

PostPosted: Tue Feb 22, 2005 1:30 pm  Back to top 
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goodyfresh741
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#7
Um, that's not completely true

jli wrote:
find all the factors of the coeficient of the one with the largest power

find the factors of the number at the end

divide them in all the possible ways and put a plus/minus in fron of the fractions and those are the possible roots


That's partially correct, but the roots you get as a result are all the possible RATIONAL roots, and some or all of the numbers you get that way may not be roots at all. The process helps not one bit in finding the irrational roots. However, on math contests, the so-called "rational zero theorem" helps ALOT, because many times, the roots of polynomials in math contests problems are integers or fractions.
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PostPosted: Tue Feb 22, 2005 8:43 pm  Back to top 
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Scrambled
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#8
Re: Polynomials

p_adic wrote:

R(x) = x^8 + 5x^5 + 5x^3 + 1 = 0 can be solved by dividing through by x^4:

x^4 + 5(x + 1/x) + 1/x^4 = 0

Let u = x + 1/x

Then u^2 = x^2 + 1/x^2 + 2,

(u^2 - 2)^2 = x^4 + 1/x^4 + 2

So x^4 + 1/x^4 = (u^2 -2)^2 -2, hence the equation is

(u^2 -2)^2 - 2 + 5u = 0

which can be solved for u by the quartic formula and then for u.

that equation is symmetric, (having the same terms on either side), and a symmetric equations can be solved this way in generla

and i dont think you need that quartic?
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PostPosted: Wed Feb 23, 2005 11:02 am  Back to top 
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ZetaX
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#9
goodyfresh741: can you please give me a link or something like that to a solution for quintic equations using Thetafunctions¿
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PostPosted: Wed Feb 23, 2005 11:33 am  Back to top 
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bénabar
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#10
Re: Unsolvability of quintic

goodyfresh741 wrote:
In the 1800's (I'm not sure exactly when) Abel proved that it's impossible to find an algebraic solution to a general equation of degree >=5.


Isn't it Galois? Of course, It is not very important... Wink

PostPosted: Wed Feb 23, 2005 11:36 am  Back to top 
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bénabar
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#11
...but Galois was french Razz Razz Razz.

PostPosted: Wed Feb 23, 2005 11:54 am  Back to top 
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ZetaX
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#12
I think Abel proofed it for rational polynomials, Galois in a more general case for fields. (But's possible that I'm wrong)
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PostPosted: Wed Feb 23, 2005 12:09 pm  Back to top 
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goodyfresh741
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#13
Abel and Galois

Yes, Abel proved it first, but Galois proved it for a more general class of polynomials over fields.
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PostPosted: Fri Feb 25, 2005 2:59 pm  Back to top 
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AntonioMainenti
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#14
This guy (http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Ruffini.html) proved that the quintic is unsolvable by radicals around 1800 and sent it to Lagrange and others but no one seemed to care or believe him. I know about him because he was a northern Italian whose last name was Ruffini. My father is a northern Italian whose last name is Ruffini Mr. Green We could be related..
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PostPosted: Fri Feb 25, 2005 7:02 pm  Back to top 
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