Suppose $f(x)$ has a factor $(x-a)$ in the field $F[X]$. Then, is $f(x)$ reducible over $F$? I know that we can write
$f(x) = (x-a)q(x)$ for some $q(x) \in F[x]$. Now, of course, deg$[(x-a)] = 1$ and deg$[f(x)] \geq 1$. This means that it is possible that deg$[f(x)] = 1$ (which would mean that deg$[q(x)] = 0$, which is fine). So, we cannot always express $f(x)$ as a product of two polynomials whose respective degrees are strictly less than the degree of $f(x)$.
The problem is that this argument seems to produce a very counter-intuitive result for me. Is the argument even correct?
Definition of a reducible polynomial:
A nonconstant polynomial $f(x) \in F[x]$ is reducible over $F$ if $f(x)$ can be expressed as a product $g(x)h(x)$ of two polynomials $g(x)$ and $h(x)$ in $F[x]$ both of lower degree than the degree of $f(x)$.
Best Answer
As you've seen, the result is not true if $\deg f = 1$.
Consider the polynomial $f(x) = x$. It has a factor, namely, itself but it is irreducible.
In fact, your title says
Even that is not true. To see this, pick your favourite irreducible polynomial $f$ (not necessarily of degree $1$). Then, $f$ is a factor of $f$ but it is not reducible.
For example, you could choose $f(x) = x^2 + 1 \in \Bbb R[x]$.
What you do have is that if $f(x) \in F[x]$ has a linear factor in $F[x]$, then $f$ is reducible iff $\deg f \ge 2.$
A proof of this has been given in your other question here.