Question 1102160
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With the question as you show it, all we know is that there is a root at x=1.<br>
With only that, the answer can't be that the multiplicity is 0, as you say you thought it should be; a multiplicity of 0 would mean there is not a root there.<br>
But with nothing else in the given information, you could certainly have a root of multiplicity 1 at x=1; y=x-1 would be the simplest example.<br>
So if the correct answer is that the root has to have multiplicity 2, then there is something missing in your statement of the problem.