Much of this would be covered in "basic ring theory". Let's say that $R$ is a commutative ring with unity (although more general settings can be accommodated, it seems like a good place to start).
As you studied "polynomials" in calculus you were asked to consider them as functions of one or more variables. In abstract algebra there is more of an emphasis on polynomials as elements of a symbolic ring extension.
The univariate (single variable) case is to define a ring $R[X]$, where we will often refer to the symbol $X$ not as a variable (which is useful if polynomials are considered as functions) but as an indeterminate, meaning that it has no specified value in connection with the base ring of coefficients $R$. In other words $X$ is a purely symbolic value, and we define polynomials in this single indeterminate as a finite sum of natural powers of $X$ (including $1\in R$ as the special case $X^0$) times coefficients from $R$.
For example a polynomial $f(X)\in R[X]$ of degree $n$ has the form:
$$ f(X) = \sum_{k=0}^n r_k X^k \; \text{ where } \; r_k\in R \; \text{ for } k=0,\ldots,n $$
So $X^2 + 2X + 1$ is polynomial from the ring $\mathbb{Z}[X]$, or from any polynomial ring $R[X]$ in which the integers $\mathbb{Z}$ are considered to be a subring of $R$.
Addition and multiplication of polynomials is done as you learned in secondary education, using distributivity and collection of "like terms".
(1) A polynomial ring is an integral domain in the same sense as any commutative ring with unity is an integral domain, namely that a product of two elements is zero only if one or the other of the two elements is zero. (I phrase it this way because you say you are already familiar with the meaning of integral domain.)
An interesting exercise is to prove that $R[X]$ is an integral domain if and only if $R$ is an integral domain. Some familiarity with induction and the degree of a polynomial is needed to prove this.
(2) The notion of a unit for polynomial rings is again the same as for all commutative rings with unity. In particular a unit is an element that has a multiplicative inverse.
Another interesting exercise is to prove that if $R$ is an integral domain, then $R[X]$ has essentially the same units as $R$, in the sense that a unit of $R[X]$ is a "constant" polynomial $C_0 \in R$ that is already invertible in $R$.
(3) An irreducible polynomial $f(X) \in R[X]$ is an element that has factors:
$$ f(X) = g(X)h(x) $$
only of the form where $g(X)$ or $h(X)$ is a unit. This is not quite the idea you described, because we don't require merely that $g(X)$ and $h(X)$ cannot both have smaller degrees than $f(X)$. For example, $f(X) = 2X + 4$ is not irreducible in $\mathbb{Z}[X]$ because $f(X) = 2\cdot(X+2)$ is a way to factor into nonunits over $\mathbb{Z}[X]$. On the other hand, if your base ring were $R = \mathbb{Q}$, the rational numbers rather than the integers, we would consider $2$ to be a unit, and thus $f(X) = 2X+4$ is irreducible in $\mathbb{Q}[X]$.
(4) Zeros of a polynomial are the same as roots of a polynomial, and for this concept we switch back to thinking of polynomials as functions. In other words, $r\in R$ is a root (zero) of polynomial $f(X)$ if and only if assigning the value $X=r$ results in polynomial evaluation to zero, $f(r) = 0$. From an abstract algebra point of view, this "polynomial evaluation" at $X=r$ amounts to a ring homomorphism from $R[X]$ into $R$. Different values of $r\in R$ will give different ring homomorphisms.
Often a ring may lack the roots (zeros) for a polynomial that we would like to have. For example $f(X) = X^2 + 1$ has no roots over the polynomial ring $\mathbb{Z}[X]$, or even over the rational or real polynomials. But it will have two distinct roots if we extend coefficients to the complex numbers $\mathbb{C}[X]$. Much of the further steps in abstract ring theory will examine the process of constructing ring extensions that produce roots.
(5) Zero divisors in polynomial rings have again the same definition as for commutative rings with unity in general. That is, if a (nonzero) element can be multiplied by another nonzero element to give a product zero, then we can say that element is a zero divisor. Notice that an integral domain is simply a commutative ring with unity that has no zero divisors.
Note that typically an author excludes zero itself from consideration as a zero divisor, but perhaps it makes sense to always qualify the meaning by saying an integral domain has no zero divisors other than zero.
We can thus restate an earlier proposed exercise as proving $R[X]$ has no zero divisors if and only if $R$ has no zero divisors.
