Solving uartics

  • #1
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Main Question or Discussion Point

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

I have the equation

[tex]x^{4}+5x^{2}+4x+5=0[/tex]

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.
 

Answers and Replies

  • #2
Mentallic
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If the quartic has "nice" quadratic factors, then they will be of the form:

[tex]x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)[/tex]

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 [itex]\pm[/itex] operator.
 
  • #3
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If the quartic has "nice" quadratic factors, then they will be of the form:

[tex]x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)[/tex]

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 [itex]\pm[/itex] operator.
Oh, I see, but how do we know when the factors are "nice"?
 
  • #4
Mentallic
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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 [itex]x^4+5x^2+4x+5[/itex] we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, [itex]x^4+6x^2+4x+5[/itex] then we find no solutions for a,b in the system of equations.
 
  • #5
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Ok, thanks!
 
  • #7
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Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link :smile:
 
  • #8
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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 [itex]x^4+5x^2+4x+5[/itex] we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, [itex]x^4+6x^2+4x+5[/itex] 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?
 
  • #9
Mentallic
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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
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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
Mentallic
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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=[itex]\pm1[/itex]
4a=4 -> a=1, b=-1



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
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For the second system, I don't get any solutions. Therefore, that can't be the right system.
 
  • #13
Mentallic
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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
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Ah, that's what I thought. Thanks!
 
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