# Solving uartics

1. Sep 23, 2012

### 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.

2. Sep 23, 2012

### Mentallic

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.

3. Sep 23, 2012

### BloodyFrozen

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

4. Sep 23, 2012

### Mentallic

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.

5. Sep 23, 2012

### BloodyFrozen

Ok, thanks!

6. Sep 23, 2012

### mathwonk

7. Sep 24, 2012

### BloodyFrozen

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

8. Sep 24, 2012

### BloodyFrozen

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

9. Sep 24, 2012

### Mentallic

I don't know, what did you get?

10. Sep 24, 2012

### 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

11. Sep 24, 2012

### Mentallic

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

For this system we have no real solution, so what does that tell you?

12. Sep 24, 2012

### BloodyFrozen

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

13. Sep 24, 2012

### Mentallic

Exactly, so what must your factors be?

14. Sep 25, 2012

### BloodyFrozen

Ah, that's what I thought. Thanks!