Let $P(x)=a_nx^n +a_{n-1}x^{n-1}+\cdots +a_1x +a_0$, where all the coefficients are integers.
The Rational Roots Theorem says that all rational roots of this polynomial have shape $\frac{p}{q}$, where $p$ divides $a_0$ and $q$ divides $a_n$. Please note that possibly $p$ is negative.
For any specific integers $a_n$ and $a_0$, this produces a finite list of candidates, and we can usually check them all without difficulty by substituting.
The polynomial need not have any rational roots, though cubics in school exercises usually do!
If our polynomial is $P(x)$, and $a$ is a root, then $x-a$ divides $P(x)$. It seems to be a bit of a waste of time to divide $P(x)$ by $x-a$ to check whether $a$ really is a root, since "plugging in" is easier, and less likely to lead to error.
However, synthetic division can be useful. Suppose for example $P(x)$ is a cubic, and we have found a rational root $a$, and only one. We can then calculate $Q(x)=\frac{P(x)}{x-a}$ and obtain a quadratic. Roots of this can then be easily found using the Quadratic Formula.
Remark: We are not dividing the polynomial by one of the zeros, we are dividing by the linear polynomial $x-a$, where $a$ is a zero of the polynomial.
I do not understand the modular arithmetic part of the question. The ideas of modular arithmetic can be extended to polynomials, even polynomials with real coefficients, and much more. So we can say that $a$ is a root if and only if $P(x)\equiv 0\pmod{x-a}$. That is just a restatement of the result that $x-a$ divides $P(x)$.
