Seth says  
ok so i was doing my math hw and i stared getting really annoyed with finding roots of cubics because i had no therums that didnt require trailanderror to some degree so i started googleing around and i think this formula might work the best of any other that i found. can anyone better at math than me verify that this does indeed work? http://planetmath.org/encyclopedia/CubicFormula.html 

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Seth says 


this is i mean does it work for ALL cases?  
 03 February, 2007 
Seth says 


*that (not "this")  
 03 February, 2007 
jared.nance says 


seth, that formula will have problems with certain coefficients where you will end up dividing by zero... not so good. check out mathworld for an excellent treatment of this problem, and many others as well... http://mathworld.wolfram.com/CubicFormula.html 

 03 February, 2007 
Seth says 


dang, i was hoping that it would be all in one formula  
 04 February, 2007 
Seth says 


im confused. everything in the mathworld eqution seems to come down to lines (5355) which seems to be the same as the equation from PlanetMath, just with S and T so that its easier to read  
 05 February, 2007 
Joe says 


I haven't worked it out, but you can just substitute S and T, and then Q and R into that final equation. That will most likely give you that insane equation listed on the other site. If so, that's pretty awesome. Now all you have to do is prove mathematicians of the past few hundred years wrong and generalize up to polynomials of order n.  
 06 February, 2007 
Seth says 


but thats what im saying jared told me the insane one on the other site didnt work because of divid by zero problems. btw, jared are those div by 0 problems in any specific equations? cause if i can solve for r1, then it can just divide the cubic by (xr1) and get a quadratic 

 06 February, 2007 
jared.nance says 


yeah, it seems that the division by zero occurs for certain values of the coefficients. not so good. the best thing to do to solve cubics is just follow cordano's procedure... and believe me, you don't even want to see the quartic formula. and the quintic, luckily, doesn't exist. it has been proven that you cannot analytically get the roots.  
 06 February, 2007 
Joe says 


I'm still not seeing where there is a problem dividing by zero. All of the coefficients go in the numerator and it's only constants in the denominator. Unless I'm missing something...  
 06 February, 2007 
(Guest) Seth ( forgot to sign in) says 


which coefficiants would that be for? 

 06 February, 2007 
Seth says 


http://www.1728.com/cubic2.htm thats a another thing i found. wyhat do u think of that one? 

 06 February, 2007 
Joe says 


Here's something interesting. The "cubic formula" as given above only works if all the roots are real. It does not work if two of them are complex. I'm not sure why, but I'll try to find out.  
 07 February, 2007 
Seth says 


which one? the 1728.com one? 

 08 February, 2007 
Joe says 


No, the first planet math one.  
 09 February, 2007 
Viczy says 


Don't know if this helps or not, but you can find the values of x for the critical points of f(x) as follows: ± I think that checks out. Hope it helps! 

 21 February, 2007 
Viczy says 


Gah, ignore tha A^ thingy. Damned no edits...  
 21 February, 2007 
Seth says 


ur math chacks out but im not sure it helps all that much cuz as u said, thats finding the critical points. i dont think that really helps in finding the roots though.  
 22 February, 2007 
Viczy says 


Ah, alright. My bad.  
 23 February, 2007 
