If I have polynomial $f(x) = (x-1)(x-1)(x-1)$ it is a cubic equation but only one root of $1$. I thought the number of roots (real and complex) equaled the degree of the polynomial? Does the equation technically have three roots?
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An element $a \in \mathbb{C}$ is called a root of multiplicity $k$ of a polynomial $p(x)$ if there is a polynomial $s(x)$ such that $s(a) \neq 0$ and $p(x) = (x − a)^ks(x)$.
The polynomial $f(x) = (x-1)^3$ has a root of multiplicity three.
A degree $n$ polynomial has precisely $n$ roots over the complex numbers, when the roots are counted with multiplicity. This is one way to state the fundamental theorem of algebra.
Michael Albanese
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