Solving x^3+a*x^2+b*x+c=0

[D4RK_G4ND4LF]

I wanted to solve an equatation of the form

x^3+a*x^2+b*x+c = 0
where a, b, c were given

and I had the following idea:

(x+d)*(x+e)*(x+f) = 0
(x^2 + d*x + e*x + d*e)*(x+f) = 0
x^3 + d*x^2 + e*x^2 + d*e*x + f*x^2 + d*f*x + e*f*x + d*e*f = 0
x^3 + (d*x^2 + e*x^2 + f*x^2) + (d*e*x + d*f*x + e*f*x) + d*e*f = 0
x^3 + (d+e+f)*x^2 + (d*e+d*f+e*f)*x + d*e*f = 0

a = d + e + f
b = d*e + d*f + e*f
c = d*e*f

So I got 3 equatations with 3 unknown variables so it should be possible to solve this
I tried to do it and after filling a few pieces of paper with formulars I surrendered
According to http://www.wolframalpha.com/input/?i=a+%3D+d%2Be%2Bf%2C+b+%3D+d*e%2Bd*f%2Be*f%2C+c+%3D+d*e*f
the equatation is rather complicated and I am too lazy to write it down

Can anyone post a simple step-by-step solution to this?

I require it for some complicated projectile movement

 

[Mac_MacNuggets]

Dude. Isn't the answer in the website? And why do you even need it?

 

[meOme]

Search google for the term "cubic equation" and you will find tons of stuff about this function.

 

[HINDYhat]

If you want to solve it over the rationals, you can just brute force solutions using Descarte's rational root theorem.

Otherwise, google it. I remember how to solve one of the form: x³ + ax = b, but none with the quadratic term. Just google it lol.

 

[D4RK_G4ND4LF]

I am not looking for a solution of a cubic equatation but for a solution to
a = d+e+f
b = d*e+d*f+e*f
c = d*e*f

Dude. Isn't the answer in the website? And why do you even need it?

there must be an easier solution than that

 

[HINDYhat]

If you're looking for the solution to that, you're essentially looking for solutions to a cubic equation, and as you've noticed, those tend to be pretty bulky unless if you make clever substitutions.

You don't always get simple solutions when you have a system of 3 equations and 3 variables (you might be thinking of a system of linear equations, and this is not that kind of a system).

There are several solutions on the wikipedia page on cubic equations.

 

[darkrider]

so far you have this

x^3 + d*x^2+e*x^2+f*x^2 + d*e*x+d*f*x+e*f*x + d*e*f = 0

which can be written like:

x^3+x^2*(d*e*f)+(x*e*f+x*d*f+x*d*e)
x^3+x^2*(d*e*f)+x*(e*f+d*f+d*e)

now we have x in every term, common factor

x(x^2+x*(d*e*f)+(e*f+d*f+d*e)) = 0

Now, if we want a multiplication to be = 0 either one of those factors have to be 0
In case x = 0 you already have a result and we have to work out this

x^2+x*(d*e*f)+(e*f+d*f+d*e)=0

so far this is getting kinda hard and I'm at work xD (If not solved 'till tonight i'll try later

EDIT:

Continue

now we have 3 terms
x^2
x*(d*e*f)
(e*f+d*f+d*e)

we can use this formula

(-b+-(b^2-4*a*c)^1/2)/2*a

which will result in

-(d*e*f)+-(def)^2-(4*1*(e*f+d*f+d*e))^1/2)/(2*1) = 0

(2*1)*0 = 0

then

-(d*e*f)+-(d*e*f)^2-(4*(e*f+d*f+d*e))^1/2) = 0

+-(d*e*f)^2-(4*(e*f+d*f+d*e))^1/2 = (d*e*f)

sqrt to pow

(d*e*f)^2)-(4*(e*f+d*f+d*e)) = (d*e*f)^2

now we have this

-4*(e*f+d*f+d*e) = 1

we have 2 terms multiplying

-4

e*f+d*f+d*e

I'm gonna try to do magic from here xD

 

[HINDYhat]

Nah you made mistakes. There is no way to solve this quickly without making clever substitutions, and clever people came up with clever substitutions which are documented on wikipedia. Just look it up, and I'd advise everyone else to stop trying, because it's pretty hard.

 

[Dr Super Good]

Take your pick....
I advise revising your formula, or try using matrix calculations.

 

[Fingolfin]

The formula would be much easier to read and understand if you'd have put proper bracketing in the third and fourth line.

 

[D4RK_G4ND4LF]

The formula would be much easier to read and understand if you'd have put proper bracketing in the third and fourth line.

thanks! this solves all my problems!

edit:
oh...wait...
thanks! this solves all my problems!

 

[Fingolfin]

I see what you did there.

 

[PurgeandFire]

I suggest you find an alternative, even if you get the solution down. The size of the equation is huge and would probably the slowest method to choose. :\