Are all quadratics factorable into a product of two binomials? I'm learning algebra in school, and my teacher said that all quadratics are factorable into a product of two binomials. I then realized however that some quadratics would have imaginary roots, and therefore wouldn't be able to be put into factored form. Who's wrong here, my teacher or me? For example, can $x^2 + 4x + 1$ even be expressed in factored form? Thanks in advance.

Let $P(z) = \sum_{i=0}^n a_i z^i$, so that $P$ is a polynomial of degree $n$. Suppose that $z,a_1,\dots,a_n \in \mathbb{C}$. Then, there exists $r_1,\dots,r_n \in \mathbb{C}$ such that $P(z) = a_n\prod_{i=1}^n (z-r_i)$. This is more or less what the Fundamental Theorem of Algebra states (or implies, depending on who you ask). Thus, under these circumstances, every polynomial of degree $n$ can be written as a product of exactly $n$ binomial factors, not necessarily distinct.
However, if we suppose that $r_1,\dots,r_n \in \mathbb{R}$, then this is no longer the case, even if we also have $z,a_1,\dots,a_n \in \mathbb{R}$. For example, the polynomial $P_0(z) = z^2+z+1$ has no such representation, and equivalently, there is no $z^* (\in \mathbb{R})$ such that $P_0(z^*) = 0$. Since $\mathbb{Q}, \mathbb{Z},$ and $\mathbb{N}$ are subsets of $\mathbb{R}$, they have similar properties.
