Let $p > 2019$ be a prime no. Consider the polynomial $f(x) = (x^2-3)(x^2-673)(x^2-2019)$ How many roots can $f$ possibly have in finite field $F_p$

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Let $p > 2019$ be a prime number. Consider the polynomial $f(x) = (x^2-3)(x^2-673)(x^2-2019)$ How many roots can $f$ possibly have in finite field $F_p$ ?

1. $0$

2. $2$

3. $3$

4. $6$

I have seen this question on this site but I could not understand it from there. So I posted this question again. Here I see that $2019 = 3×673$. Also $F_p = \{0,1,2,3,...,p-1\}$, Here $3,673,2019 \in \{0,1,2,3,...,p-1\}$ for $p >2019$. So options 2,3 can be possible. But correct options are $2,4$ and I could not approach.

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