I am reading Saito's Fermat's Last Theorem: Basic tools and on page 15, it was claimed that
Proposition 1.4. Let $K$ be a field with char($K$) $\neq 2$, and let E be an elliptic curve over $K$. Then all $2$-torsion points of $E$ are $K$ rational if and only if there exist distinct nonzero elements $n$ and $m$ in $K$ such that $E$ is isomorphic to the elliptic curve defined by $$y^2 = x(x-n)(x-m)$$
Proof. Suppose $E$ is defined by $y^2=f(x)$. Let $L$ be an extension of $K$. We prove that a point $P=(s,t)\neq O$ in $E(L)$ is of order $2$ if and only if the $f(s)=t=0$. By the definition of the group law, $P$ is of order $2$ iff the tangent line at $P$ passes through $O=(0:1:0)$. Since a line in $\mathbf P^2$ passesthrough $O$ iff it is parallel to the $y$-axis, the tangent line must be the line $x=s$. This is equivalent to the fact that the system of eqeuation $x=s, y^2 = f(x)$ has a multiple root at $(x,y)= (s,t)$.
This, in turn, is equivalent to $f(s) =t=0$.
Then the author claimed that
Thus, all the $2$-torsion are $K$-rational if and only if $f(x)$ is decomposed into linear factors over $K$.
Does the author somehow assume there are at least $3$ distinct $2$-torsion points on $E$, which are all $K$-rational?
EDIT: I somehow get that I need to use the definition of elliptic curves where $f(x)$ is a cubic polynomial, and by definition has no multiple roots. The problem is solved now.