I'm reading the section about field extensions in Dummit and Foote. It says that if you have a field $\mathbb{F}$ and a polynomial $p(x) \in \mathbb{F}[X]$ we can ask the question whether there exists an extension $\mathbb{K}$ of $\mathbb{F}$ containing a solution to $p(x) = 0$. Then it says that "we may assume here that the polynomial $p(x)$ is irreducible in $\mathbb{F}[X]$ since a root of any factor any factor $p(x)$ is certainly a root of $p(x)$ itself.
I'm having trouble understanding why you can assume that $p(x)$ is always irreducible because I think you can have reducible polynomials that don't have any roots over your field.
For example if we have $p(x) = (x^2 +1)(x^2 +3)$ and our polynomial ring is $\mathbb{R}[X]$ then $p(x)$ doesn't have any roots over $\mathbb{R}$ but it is reducible.
As pointed out in the comments, if $\alpha$ is a root of $p(x)$, then it must be a root of one of its irreducible factors since we are working over a field. If $p(x) = \pi_1(x)^{e_1} \cdots \pi_t(x)^{e_t}$ is the prime factorization of $p(x)$ and $\alpha$ is a root of $p(x)$, then $$ 0 = p(\alpha) = \pi_1(\alpha)^{e_1} \cdots \pi_t(\alpha)^{e_t} \, . $$ Since we are working over a field (which is in particular an integral domain), then we must have $\pi_i(\alpha) = 0$ for some $i$. For instance in your example, if $0 = p(\alpha) = (\alpha^2 + 1)(\alpha^2 + 3)$, then either $\alpha^2 + 1 = 0$ or $\alpha^2 + 3 = 0$, i.e., $\alpha$ is either a root of the irreducible factor $x^2 + 1$ or the irreducible factor $x^2 + 3$.
The usual strategy for constructing a field extension containing a root of $p(x)$ also requires $p$ to be irreducible. If you look just below at the proof of Theorem 3 in $\S13.1$ (p. 512), you'll see that this field extension is defined as $F[x]/(p(x))$. Since $F[x]$ is a PID and $p(x)$ is irreducible, then $(p(x))$ is a maximal ideal so this quotient is indeed a field. On the other hand, quotienting out by the ideal generated by a reducible polynomial will not produce a field since it will contain zero-divisors.
One certainly can quotient by a reducible polynomial and still obtain an interesting object. If the polynomial is at least separable, the resulting quotient is called an étale algebra.
Edit: To respond to your question in the comments, it is certainly possible that adjoining the roots of one irreducible factor of $p(x)$ will not produce a field containing all roots of $p(x)$. However, we can adjoin the roots of the irreducible factors successively.
For example, consider the polynomial $p(x) = (x^2 - 2)(x^2 - 3) \in \mathbb{Q}[x]$. Adjoining the roots of the first irreducible factor produces the field $F := \mathbb{Q}(\sqrt{2}) = \frac{\mathbb{Q}[x]}{(x^2 - 2)}$, but one can show that $F$ does not contain the roots of $x^2 - 3$. So we repeat the process and adjoin these roots to $F$, obtaining $K := F(\sqrt{3}) = \frac{F[x]}{(x^2 - 3)} = \mathbb{Q}(\sqrt{2},\sqrt{3})$.