Fundamental theorem of algebra for finite fields
Multivariate polynomials over an infinite field can have infinitely many roots, as pointed out by other users.
As for the univariate case, the answer is yes: if $f$ is a (univariate) polynomial over a field $K$ and $a\in K$ is a root, then we can use the division algorithm to show that $f(x)=(xa)q(x)$ for some polynomial $q$ over $K$.
Note that this isn't the fundamental theorem of algebra, which says that every complex (univariate) polynomial of degree $n$ has exactly $n$ roots, counting multiplicity, in the complex numbers.
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Comments

Jessica 6 months
Over the real numbers, a polynomial with degree $n$ has at most $n$ roots, regardless of whether it's a single variate or a multivariate polynomial. Is the same true when the polynomial is over a finite field like $\mathbb{Z}_p$?

Seth over 8 yearsMultivariable polynomials of degree $n$ don't have at most $n$ roots. They generally have infinitely many roots.

Matt E over 8 yearsDear eeeeeeeeee, The polynomial $x^2 + y ^2  1$ is a $2$variable polynomial of degree $2$, which certainly has more than $2$ roots over $\mathbb R$. So I think there is an error in your assertion. Regards,

Jessica over 8 yearsAw, shucks. Thanks for the correction.

rschwieb over 8 yearsBut finite fields do have algebraic closures, where any polynomial will split into linear factors.

Jessica over 8 years@rschwieb even multivariable polynomials?

rschwieb over 8 years@eeeeeeeeee No, not multivariable polynomials. The statement of the fundamental theorem of algebra has nothing to do with multivariable polynomials, so it did not occur to me to mention them.
