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

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- Apr 25th 2010, 02:44 PMsuperdudeIs there a cubic equation like the quadratic equation?
For example if I had something like $\displaystyle (\lambda)(\lambda^2-1)-(\lambda-1)+(1-\lambda)$ how would I find the roots? Is there away to quickly factor it or an equation similar to the quadratic equation?

- Apr 25th 2010, 02:56 PMskeeter
$\displaystyle (\lambda)(\lambda^2-1)-(\lambda-1)+(1-\lambda) = 0$

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

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

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

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

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

$\displaystyle (\lambda-1)^2(\lambda+2) = 0$ - Apr 25th 2010, 03:32 PMmaddas
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. - Apr 25th 2010, 03:53 PMDeadstar