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.
