If we try to find the roots of $x^3-27,$ and we come up with $x=\sqrt[3]{27}=3,$ it is possible to contemplate the idea of having a root $(3)$ with mutiplicity $3.$ After all, the fundamental theorem of algebra calls for a maximum of 3 real roots – the alternative being 1 real and 2 complex.
Clearly, $(x-3)^3 \neq x^3-27,$ and the other two roots are complex: $-3/2i(\sqrt 3 – i)$ and its complex conjugate, $3/2 i (\sqrt 3 + i).$
Probably, if instead of just looking into $\sqrt[3]{27},$ I had gone directly to the cubic formula, I would have ended up with negatives under the roots. But the cubic formula is scary.
So the question is, how could I have guessed that instead of 1 real root with multiplicity of 3, I was going to end up with 1 real root and 2 complex roots?
Best Answer
In general you can rely on the following to simplify your life a bit:
Therefore you have a very nice way to check wether it has multiple roots or not: factorize it in a convenient field, for example in $\mathbb{Q}$ where you can use criterions such as Eisenstein and so on. And once you have the factorization you should check that the factors do not have roots in common. This reduces at least the problem to a simpler one.
But in some cases you can already guess from the looks of the polynomial, as you said. In this case for example:
The reason: if $a>0$ then $\sqrt[N]{a}>0$ is a root already in $\mathbb{R}$. But now let $\zeta^{k}\in \mathbb{C}$ be the $N$ different $N^{th}$-roots of unity in $\mathbb{C}$. You know they are different because you can write them down explicitly with the exponential form of complex numbers for example, and you know those are all there are because of the fundamental theorem of algebra. Then, using that the powers of the product is the product of the powers and again arguing with the exponential form of complex numbers and the fundamental theorem of algebra, you see that the $N$ different roots of $x^{N}-a$ are $$ \zeta^{k}\sqrt[N]{a}$$ for $k=0,1,\ldots,N-1$.