Why are there 3 roots to a cubic equation?

okkvlt

I find complex numbers very fascinating. But i dont understand something.
Why does a cubic equation have 3 answers instead of 6?

I know that there are 3 cube roots of a complex number, and the imaginary part of the complex number can be either positive or negative, so there should be 6 answers. Actually, if each side of the formula was independent of the other, there would be 3*4=12 answers.

I tried reading an article on the internet about galois theory, but it used alot of jargon about fields and groups that i didnt understand.

Could somebody explain to me why there are only 3 roots, without all the jargon?

matt grime

It is nought to do with Galois theory.

What you've written implies that you think that a+ib and a-ib cube to give the same number. That is clearly not true: just try it.

All you need to use is the euclidean algorithm for polynomials: if I have a polynomial f(x) and f(a)=0, then I can write f(x)=(x-a)g(x) where the degree of g(x) is one less than the degree of f(x).

Finally, why have you chosen cubic equations? Do you accept that quadratics have two roots, and if so why have you accepted that without question?

HallsofIvy

NO. "the imaginary part of the complex number can be either positive or negative" is true only for square roots of real numbers.

In the complex numbers any n^th degree polynomial can be written as the product of exactly n linear terms. In particular, any cubic polynomial can be written as the product of 3 linear terms. Setting each term equal to 0 will give at most 3 distince solutions.

tiny-tim

Hi okkvlt!

You're obviously thinking that the conjugate of a real cubic equation is itself, and therefore the conjugates of its roots must also be roots.

That's correct!

But that doesn't mean that there are 6 answers … one root of a real cubic equation will always be real, and the other two will be conjugates of each other (whether real or not) …

so including the conjugates still leaves the total as three!