Solving uartics

BloodyFrozen
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.

Homework Helper
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.

BloodyFrozen
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"?

Homework Helper
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.

BloodyFrozen
Ok, thanks!

BloodyFrozen
Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link

BloodyFrozen
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?

Homework Helper
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?

BloodyFrozen
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

Homework Helper
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

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?

BloodyFrozen
For the second system, I don't get any solutions. Therefore, that can't be the right system.

Homework Helper
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?

BloodyFrozen
Ah, that's what I thought. Thanks!