Let $f(x) \in F[x]$ where $F$ is a field. The remainder theorem states that the remainder when $f(x)$ is divided by $x-a$ is $f(a)$. This implies that $f(a) = 0$ if and only if $x-a$ is a factor. This means that there is no irreducible polynomial $p(x) \in F[x]$ other than $x-a$ up to a constant such that $p(a) = 0$ since that implies that $x-a$ must be a factor of $p(x)$. Since this can be said about every $a \in F$, this must imply that there are no irreducible polynomials of $\deg > 1 $ with roots in $F$ although the converse might not be true. So Therefore, we have irreducible (of degree greater than 1) implies no roots. The contrapositive of this statement would be roots implies reducible. However, one of the theorems in the text that I'm reading states the following:
Let $F$ be a field. If $f(x) \in F[x]$ and $\deg f(x)$ is $2$ or $3$, then $f(x)$ is reducible over $F$ if and only if $f(x)$ has a zero in $F$.
Following the logic above, shouldn't the more general statement be roots implies reducible? However, I haven't seen it mentioned anywhere. Is the theorem a special case where we can say if and only if or is there anything wrong with my reasoning?
If $f(x)$ (of degree greater than one) has roots in $F$ it is indeed reducible as you said, but for general $f(x)$, it may be reducible without having any roots in $F$. For example, $f(x) = x^4+2x^2+1 = (x^2+1)^2$ is clearly reducible over $\mathbb{Q}$ but has no roots in $\mathbb{Q}$.
But the converse holds if the degree of $f(x)$ is $2$ or $3$, because if such $f$ is reducible, one of its factors must be of degree $1$, and as you said, it would have a root.