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 10 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 about 9 yearsMultivariable polynomials of degree $n$ don't have at most $n$ roots. They generally have infinitely many roots.

Matt E about 9 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 about 9 yearsAw, shucks. Thanks for the correction.

rschwieb about 9 yearsBut finite fields do have algebraic closures, where any polynomial will split into linear factors.

Jessica about 9 years@rschwieb even multivariable polynomials?

rschwieb about 9 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.
