Dummit and Foote Abstract Algebra page 512
Given any field F and any polynomial $p(x)\in F[x]$ one can ask a similar question: does there exist an extension K of F containing a solution of the equation $p(x)=0$ Note we may assume here that the polynomial p(x) is irreducible in F[x] since a root of any factor of p(x) is certainly a root of p(x) itself
I am having trouble with the bold part. I understand that a root of a factor will also be a root of the polynomial, but I don't understand why we can assume p(x) is irreducible.
Any explanation of this would be great. Thanks.
A field containing a root of an irreducible factor of $p(x)$ is a field containing a root of $p(x)$, as desired. Therefore it suffices to solve the problem for the case of $p(x)$ irreducible.
Generally, if $\,S\subset F[x]^*\,$ is closed under multiplication then $\,S= F[x]^*\!\iff S$ contains every irreducible (and every unit), i.e. the monoid $\,F[x]^*$ is generated by its irreducibles (and units).
In your case $S$ is the set of polynomials having a root in some extension field, which is clearly closed under multiplication since if $f(x)$ has a root in $K$ so too does $f(x)g(x)$.