Consider a cubic equation:
$(x-2)(x-2)(x-1)=0$.
Does it have $3$ integer roots (where $ 2 $ are coincident) or $2$ integer roots? Or are both statements correct?
$\endgroup$1 Answer
$\begingroup$The set of roots is $\{1,2\}$. Therefore, it has two roots.
However $1$ is a simple root, whereas $2$ is a double root. So, if we count the roots with their multiplicities (that is, a simple root counts as one root, a double root as two roots and so on), it has $3$ roots, as every cubic does.
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