Is there a cubic equation like the quadratic equation?

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• April 25th 2010, 03:44 PM
superdude
Is there a cubic equation like the quadratic equation?
For example if I had something like $(\lambda)(\lambda^2-1)-(\lambda-1)+(1-\lambda)$ how would I find the roots? Is there away to quickly factor it or an equation similar to the quadratic equation?
• April 25th 2010, 03:56 PM
skeeter
Quote:

Originally Posted by superdude
For example if I had something like $(\lambda)(\lambda^2-1)-(\lambda-1)+(1-\lambda)$ how would I find the roots? Is there away to quickly factor it or an equation similar to the quadratic equation?

$(\lambda)(\lambda^2-1)-(\lambda-1)+(1-\lambda) = 0$

$(\lambda)(\lambda^2-1)-(\lambda-1)-(\lambda-1) = 0$

$(\lambda)(\lambda+1)(\lambda-1)-2(\lambda-1) = 0$

$(\lambda-1)[\lambda(\lambda+1) - 2] = 0$

$(\lambda-1)(\lambda^2 + \lambda - 2) = 0$

$(\lambda-1)(\lambda-1)(\lambda+2) = 0$

$(\lambda-1)^2(\lambda+2) = 0$
• April 25th 2010, 04:32 PM
maddas
Don't ask about the cubic equation. You don't want to see it.
Don't ask about the quartic equation. It's even worse.
Really don't ask about the quintic equation. It requires magic darker and eviler than radicals to solve such elusive beasts.

If you must, the secret lies here. But its dangerous to go alone. Take this.
• April 25th 2010, 04:53 PM
Deadstar
Quote:

Originally Posted by maddas
Don't ask about the cubic equation. You don't want to see it.
Don't ask about the quartic equation. It's even worse.
Really don't ask about the quintic equation. It requires magic darker and eviler than radicals to solve such elusive beasts.

If you must, the secret lies here. But its dangerous to go alone. Take this.

Holy **** at the quintic one. That's some mind blowing stuff.

@OP, just use Newton's method.

Guess a solution, see if it gives you an answer close to 0. Stick that in Newton's method and iterate.