Finding Quartic Roots Without Knowing One Factor

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Discussion Overview

The discussion revolves around methods for finding the roots of quartic equations, specifically the equation x4 + 5x2 + 4x + 5 = 0. Participants explore alternative approaches to Ferrari's formula and the conditions under which certain factorization methods may be applicable.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant inquires about methods to find quartic roots without knowing one factor, suggesting a known factor might be x2 + x + 1.
  • Another participant proposes that if the quartic has "nice" quadratic factors, they could be expressed in a specific form, leading to a system of equations to solve for coefficients.
  • There is a discussion about the uncertainty of when factors can be considered "nice," with one participant stating that it is not always clear and that checking for rational roots is necessary.
  • Participants discuss the outcomes of their attempts to solve systems of equations derived from the quartic, noting that one system yields no solutions, raising questions about the validity of their approach.
  • One participant shares a link to a method attributed to Euler, which is acknowledged as lengthy but interesting.
  • Another participant expresses confusion regarding the systems of equations they derived and whether they proceeded correctly, leading to further clarification and discussion about the implications of having no solutions in one system.

Areas of Agreement / Disagreement

Participants express uncertainty about the conditions under which quartic factors are "nice" and whether their derived systems of equations are correct. There is no consensus on a definitive method for finding quartic roots without prior knowledge of one factor.

Contextual Notes

Participants mention specific quartic equations and their attempts to solve them, indicating that the nature of the coefficients and the structure of the equations can significantly affect the solvability of the systems derived from them.

BloodyFrozen
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Is there a method to finding the roots of quartics besides Ferrari's formula?

I have the equation

x^{4}+5x^{2}+4x+5=0

I know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?

Thanks.

My bad about the title. It's supposed to be Quartics instead of uartics.
 
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If the quartic has "nice" quadratic factors, then they will be of the form:

x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)

Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the \pm operator.
 
Mentallic said:
If the quartic has "nice" quadratic factors, then they will be of the form:

x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)

Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the \pm operator.

Oh, I see, but how do we know when the factors are "nice"?
 
BloodyFrozen said:
Oh, I see, but how do we know when the factors are "nice"?

We don't. We can always check to see if they are, just like we check to see if there are any rational roots.

With the quartic x^4+5x^2+4x+5 we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, x^4+6x^2+4x+5 then we find no solutions for a,b in the system of equations.
 
Ok, thanks!
 
Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link :smile:
 
Mentallic said:
We don't. We can always check to see if they are, just like we check to see if there are any rational roots.

With the quartic x^4+5x^2+4x+5 we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, x^4+6x^2+4x+5 then we find no solutions for a,b in the system of equations.

I end up getting two systems but one of them has no solution. Did I proceed correctly?
 
BloodyFrozen said:
I end up getting two systems but one of them has no solution. Did I proceed correctly?

I don't know, what did you get?
 
  • #10
Sorry for not having LATEX since I'm on my phone, but I got:

System 1:

a+b=0
5+1+ab=5
5a+b=4

System 2:

a+b=0
-5-1+ab=5
-5a-b=4
 
  • #11
BloodyFrozen said:
Sorry for not having LATEX since I'm on my phone, but I got:

System 1:

a+b=0
5+1+ab=5
5a+b=4
Right, so simplifying this system, we have

a+b=0
ab=-1
5a+b=4

Which we can then deduce,
a=-b
therefore,
a2=1 -> a=\pm1
4a=4 -> a=1, b=-1
BloodyFrozen said:
System 2:

a+b=0
-5-1+ab=5
-5a-b=4
For this system we have no real solution, so what does that tell you?
 
  • #12
For the second system, I don't get any solutions. Therefore, that can't be the right system.
 
  • #13
BloodyFrozen said:
For the second system, I don't get any solutions. Therefore, that can't be the right system.

Exactly, so what must your factors be?
 
  • #14
Ah, that's what I thought. Thanks!
 

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