Here is some further context for the notion of an algebraically closed field, connecting it with irreducible polynomials and finite-degree field extensions.

The first of these is probably familiar at least in the case of polynomials over Q. A polynomial $P \in F[T]$ is said to be irreducible (over the field $F$) if $P$ has positive degree and any factorization $P = P_1 P_2$ with $P_1,P_2\in F[T]$ has either $P_1 \in F^*$ or $P_2 \in F^*$ — that is, $P_1$ or $P_2$ is in the group of units (invertible elements) of $F[T]$. Irreducible polynomials are analogous to prime integers; in particular, note that the positive-degree condition means that the units themselves are not considered “irreducible”, which is a useful convention for the same reason that we do not regard $\pm 1$ as prime numbers.

As with primes, irreducible polynomials are characterized by the property that for $M,N \in F[T]$ the product $MN$ is a multiple of $P$ iff either $M$ or $N$ is a multiple of $P$. One direction is clear: if $P$ is reducible then the factorization $P = P_1 P_2$ (with neither $P_1$ nor $P_2$ a multiple of $P$) yields the counterexample $(M,N) = (P_1,P_2)$. The other direction follows from division with remainder. Again you’ve probably seen the corresponding argument over the integers, phrased either in terms of the Euclidean algorithm or the fact that Z is a principal-ideal domain. We take the latter route here. If $I \subseteq F[T]$ is an ideal other than $\{0\}$ then it contains a nonzero polynomial $A$ of minimal degree. Then $I$ contains all multiples of $A$, and we claim that this is all of $I$. Indeed, any polynomial $Z \in I$ can be written as $AQ+R$ for some $Q,R \in F[T]$ with $\deg R < \deg A$, and then $I$ contains $Z-AQ = R$, so $R=0$ and $Z=AQ$. Now if $P$ is irreducible, and $M \in F[T]$ is not a multiple of $P$, then the ideal $I = \{ MX + PY : X,Y \in F[T] \}$ (i.e. the ideal generated by $M,P$) contains $P$, and is principal, so $I$ has some generator $A$ of which $M,P$ are both multiples; but $P$ is irreducible, so $A$ is constant, and $I$ is all of $F[T]$. In particular, $1 = MX+PY$ for some $X,Y \in F[T]$. But then $N = N(MX+PY) = (NM)X + PY$ is a multiple of $P$, as claimed. (We could also obtain $1 = MX+PY$ using the Euclidean algorithm. The conclusion that $P|N$ also follows from unique factorization in $F[T]$, but unique factorization must in turn be proved using either the Euclidean algorithm or the principal-ideal argument.)

What has all this to do with algebraic closure? Well, a polynomial of degree $1$ is automatically irreducible, because in a factorization one or the other factor must have degree zero.
Claim: $F$ is algebraically closed if and only if every irreducible polynomial in $F[T]$ has degree $1.$
Proof: $\Rightarrow$ If $F$ is algebraically closed then any $P \in F[T]$ has a degree-1 factor $T-t_1$; if $P$ is irreducible then the complementary factor must have degree zero, so $\deg P = 1$.
$\Leftarrow$ By induction over $\deg A$, every nonconstant polynomial $A \in F[T]$ has a factorization $A = \prod_{i=1}^m P_i$ where $m>0$ and each $P_i$ is irreducible. Writing each $P_i$ as $a_i (t-t_i)$ gives $A = a \prod_{i=1}^m (t-t_i)$ with $a = \prod_{i=1}^m a_i$. So indeed $A$ factors completely.

A field extension of a field $F$ is some other $F'$ that contains $F$ (with the same field operations, so in particular $0_F = 0_{F'}$ and $1_F = 1_{F'}).$ We observed already that $F'$ is then a vector space over $F$. The extension $F'/F$ is said to be finite if $F'$ is finite-dimensional as a vector space over~$F$, in which case $\dim_F F'$ is called the degree of the extension and also denoted by $[F':F]$. The familiar example is $F=\bf R$, $F'=\bf C$, with $[{\bf C} : {\bf R}] = 2$. This is connected with irreducible polynomials as follows (which also explains the “degree” terminology): if $P \in F[T]$ is irreducible then $F' := F[T] \big/ P F[T]$ is a field extension of degree equal to $\deg P$. (This generalizes the familiar construction of $\bf C$ as ${\bf R}[i]$ modulo the relation $i^2 + 1 = 0$; here we in effect set $P(T) = 0$.) We already know that $\dim_F^{\phantom0} F' = \deg P$, and the main thing to check is that when $P$ is irreducible $F'$ is not just a commutative $F$-algebra (i.e., a ring containing a copy of $F$, and thus constituting an $F$-vector space) but even a field. That is, if $M \in F[T]$ is not a multiple of $P$ then it is invertible $\bmod P$: there exists $N \in F[T]$ such that $MN \equiv 1 \bmod P$. The key is to observe that the map $X \mapsto MX$ from $F'$ to $F'$ is an $F$-linear transformation from a finite-dimensional vector space to itself, and is thus surjective iff it is injective. [This replaces the familiar argument modulo a prime $p$, where we use the fact that a map from the finite set ${\bf Z}/p{\bf Z}$ is surjective iff it is injective.] But injectivity of $X \mapsto MX$ comes down to what we showed above: if $N \bmod P$ is in the kernel then $MN \equiv 0 \bmod P$, and since $M$ is not a multiple of $P$ we deduce that $N \equiv 0 \bmod P$ and thus represents the zero element of $F'$.

In particular, if $F$ is not algebraically closed then it has a finite extension of degree $\gt 1$. Remarkably the converse is true: if $F$ is algebraically closed then it has no finite extensions other than $F$ itself. (This is thus our fourth equivalent characterization of algebraically closed fields.) Indeed if $F'/F$ is a finite extension, say of degree $d$, then for every $x \in F'$ the $d+1$ field elements $1,x,x^2,\ldots,x^d$ have a linear dependence over $F$, so $P(x) = 0$ for some polynomial $P \in F[T]$ which has degree at most $d$ and is not identically zero. Clearly $P$ is not constant, so we can write $P(x) = a \prod_{i=1}^m (x-t_i)$ for some positive $m \leq d$ and some $t_i \in F$ and $a \in F^*$. Since $F'$ is a field, one of those $m+1$ factors must vanish, and in cannot be $a$. Therefore $x-a_i = 0$ for some $i$, whence $x = a_i \in F$. This proves that $F' = F$, QED.