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I want to know how one would go about solving an unfactorable cubic. I know how to factor cubics to solve them, but I do not know what to do if I cannot factor it. For example, if I have to solve for $x$ in the cubic equation: $$2x^3+6x^2-x+4=0$$ how would I do it?

Edit: I have heard people telling me to convert it into a depressed cubic (where the $x^2$ term disappears), but I have no idea how to do that.

Edit 2: I am aware that there is a cubic formula, that for any cubic equation $ax^3+bx^2+cx+d$, it's roots are: $$x = \sqrt[3]{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2} - \dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3} + \dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} + \sqrt[3]{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right) + \sqrt{\left(\dfrac{-b^3}{27a^3}+\dfrac{bc}{6a^2}-\dfrac{d}{2a}\right)^2 - \left(\dfrac{c}{3a}-\dfrac{b^2}{9a^2}\right)^3}} - \dfrac{b}{3a}$$ This formula is way too complicated so I do not even bother memorizing it or using it.

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Wikipedia has some info on it. –  Arthur Mar 15 at 22:55
    
However, see en.wikipedia.org/wiki/Casus_irreducibilis for a caution; if you are asked for something with three irrational roots, Cardano's gives you something formally correct but from which the real parts cannot be extracted. –  Will Jagy Mar 16 at 1:27

2 Answers 2

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Observe that $2x^3+6x^2 = 2(x^3+3x^2) = 2(x+1)^3 - 6x-2$. Now let $X=x+1$, and divising by 3, your equation is now something like $X^3+pX+t=0$, which can be solved by Cardano method.

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To depress a cubic means to write it in the form $y^3+py+q=0$ by performing a convenient substitution. It is not hard, and I'll give you a hint on how to do it yourself.

Hint. Show that every cubic equation of the form $x^3+ax^2+bx+c=0$ can be written as $y^3+py+q=0$ by performing a substitution $x=y-w$. Expand the binomials and find $w$ by letting the coefficient of $y^2$ be zero.

Once you depress a cubic, you have to solve the simpler equation $y^3+py+q=0$. The solution is given by Cardano's formula, a way simpler than the one you stated: $$y=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}.$$ In fact, if you write the equation as $z^3+3zm=2n$, the formula is even simpler: $$z=\sqrt[3]{n+\sqrt{n^2+m^3}}+\sqrt[3]{n-\sqrt{n^2+m^3}}.$$ It is not easy to come up with this, but you can see a somewhat simple proof here. This equation motivated the first appearance of complex numbers and has a very curious history, so make sure you read it!

There is a very acessible historical reading in the first chapter of Needham's book on complex analysis. Highly recommended.

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