Can someone show me how to solve cubic equations?

Question

And don't just give me a link to a site like mathworld.wolfram.com, I want you to explain it.

Answer 1

First, I will say that all the gory details for the following are listed at wikipedia.org. I'm not going to repeat them, but try to explain some of the missing steps.

Given a general cubic equation x^3+ax^2+bx+c=0, we can reduce the equation to the form

y^3+py+q=0,

where y=x+(a/3) and where p and q are somewhat complicated expressions in a,b,c (not horrible, but not worth writing out here.) This is so because, with x=y-(a/3),

(y-(a/3))^3 + a*(y-(a/3))^2 + b(y-(a/3) + c
=y^3 - a*y2 + a* y - (a/3)^3 + a* y^2 + a bunch of other stuff

and so the terms in y^2 will cancel. Work by some Italian mathematicians in the 16th century led them to believe that this was a good form into which to put the equation, because they could then solve it.

Now, try substituting the expression v-u into the reduced equation. Why? Possibly some trial and error led the original discoverers to this notion. What does happen is that when you make this substitution, you get the polynomial

v^3 - 3v^2u + 3vu^2 - u^3 + pv - pu + q = 0

Regroup a bit to get

v^3 - u^3 + (v-u)*(p-3uv) + q=0.

Then we could select v^3 - u^3 + q=0 and p-3uv=0. This is a set of simultaneous equations which we can, in principle, solve for a set of values (u,v). One way to do so is by substitution; write u=p/(3v) and substitute into the other equation. This yields a quadratic equation in v^3 which is easy to solve, and which in turn can be solved for v. This gives you the value of v-u, which gives you a solution to the reduced equation. Using the relation y=x-(a/3) gives you a solution to the original equation. I'm not writing out all the gory details because you can get all of those from a site like wikipedia or mathworld.wolfram.com.

Having said all that, it's unlikely you'd ever need to use this for anything. One useful fact is that ANY cubic equation with real coefficients has at least one real solution (thanks to the intermediate value theorem) so knowing this root (or a good approximation) will allow you to reduce your problem to a quadratic equation, which is easy to solve.

And yes, I'll include a link to a site which does list the gory details.

Answer 2

Actually, the first responsent is wrong -- there is a "one size fits all" solution. It was discovered by Tartaglia in the 1500s. The link below gives you all the details.

There is also a general formula for solving quartic equations. I believe it was discovered by Legendre around 1800 or so.

Nicholas Abel showed that there is no general formula to solve polynomials of degree five or higher. Galois developed a method for determining which equations can be solved algebraicly and which cannot.

Answer 3

Is it factorable?
(a +1)^3 then the 3 roots are a+1 = 0 or a = -1

(a+1)*(a-1)* (a+1) = (a+1)*( a2-1) = a3 +a2 +a -a -1

the roots are: a+1 = 0 or a =-1
a-1 =0 or a =1

There are imaginary solutions to some as well. Is this helpful?

Answer 4

It depends on the type of cubic equation that you have. It's not a one size fits all answer.

Some you could solve with factoring, others could be solved by logarithms, sorry it's complicated.

Answer 5

there are many ways..you can just use factor theorem to find one root..then it reduces to quadratic and can be solved..
another way is to use successive iteration in ur calculator..you need to make sure it converges..thats a great way to get a quick answer..
then of course there are numerical methods like newton raphson..
you tell me a equation and i can tell how u do..

like if you want to solve x^3+2x^2-15+x=0
then write it like this: x= (-x+15-2x^2)^1/3..
then start with an initial guess in ur calci..and take that as answer..and type this operation where x is your answer as initial guess..
press enter as many times..u can until u see answers r not changing..here i get 1.84..once you have this get others