How do you find all the zeros of $y=x^2-3x-4$ with its multiplicities?

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Answer 1 · 2016-08-08T22:01:26

This quadratic has zeros $x=4$ and $x=-1$ , both with multiplicity $1$ .

$y=x^2-3x-4=(x-4)(x+1)$

Hence zeros $x=4$ and $x=-1$ , both with multiplicity $1$ .

How do you find all the zeros of y=x^2-3x-4y=x^2-3x-4 with its multiplicities?