Can any Polynomial be factored into the product of Linear expressions?

Specifically I am wondering if...

Given a Polynomial of n degree in one variable with coefficients from the Reals.

Will every Polynomial of this form be able to be factored into a product of n linear (first degree) Polynomials, with the coefficients of these factors not being constrained to $\mathbb{R}$ but to $\mathbb{C}$ instead.

By a product of linear polynomials I mean something of the form:

$$(Ax+a)(Bx+b)(Cx+c)...$$

I am also interested in the general behaviour of polynomials if we also play around with the parameters of my question. Such as factoring polynomials with complex coefficients, or coefficients from any set for that matter. As well Polynomials in any (Positive Integer?) number of variables.

The Fundamental Theorem of Algebra states precisely that:

Fundamental Theorem of Algebra. Every nonzero polynomial $p(x)$ with coefficients in $\mathbb{C}$ can be factored, in essentially a unique way, as a product of a constant and linear terms, in the form $$p(x) = a(x-r_1)\cdots(x-r_n)$$ where $a$ is the leading coefficient of $p(x)$ and $n$ is the degree of $p(x)$.

This result does not hold in general. In some cases, you cannot factor them (you cannot in $\mathbb{R}$, for instance, where $x^2+1$ cannot be written as a product of linear terms; over $\mathbb{Q}$, there are polynomials of arbitrarily high degree that cannot be factored at all, let alone into a product of linear terms). In some cases, the factorization is not unique: in the quaternions, the polnomial $x^2+1$ can be factored in infinitely many essentially distinct ways as a product of linear terms, e.g., $x^2+1 = (x+i)(x-i) = (x+j)(x-j) = (x+k)(x-j) = \cdots$.

A field $F$ is said to be algebraically closed if and only if every nonconstant polynomial $p(x)$ with coefficients in $F$ has at least one root. It is then easy to verify that $F$ is algebraically closed if and only if every nonzero polynomial can be factored into a product of linear terms, as above. Subject to some technical assumptions (The Axiom of Choice), every field is contained in an algebraically closed field, and for every field there is a "smallest" algebraically closed field that contains it, so there is a "smallest" larger field $K$ where you can guarantee that every polynomial factors. This does not hold for arbitrary sets of coefficients (e.g., the quaternions, because they are non-commutative; or rings with zero divisors; and others).

Once you go beyond one variable, it is no longer true that a polynomial can always be factored into terms of degree one, even in $\mathbb{C}$. For example, the polynomial $xy-1$ in $\mathbb{C}[x,y]$ cannot be written as a product of polynomials of degree $1$, $(ax+by+c)(rx+sy+d)$. If you could, then $ar=bs=0$, and we cannot have $a=b=0$ or $r=s=0$, so without loss of generality we would have $b=0$ and $r=0$, so we would have $(ax+c)(sy+d) = xy-1$. Then $ad=0$, so $d=0$ (since $a\neq 0)$, but then we cannot have $cd=-1$. So no such factorization is possible.