Prove that the ideal $\langle 2, X\rangle$ in $\mathbb{Z}[X]$ is not principal

**My attempt** : I found the answer here

In the proof of the theorem it is written that

suppose that the ideal $I=\langle2,X\rangle$ in $\mathbb{Z}[X]$ is principal.Then there exist $f(X) \in \mathbb{Z}[X]$ such that $I=\langle f(X)\rangle$. As $2 \in I$ there exist $g(x) \in \mathbb{Z}[X]$ such that $2= f(X)g(X)$

Hence $\deg f(X) + \deg g(X) =\deg f(X)g(X) =\deg 2 =0$

This $f(X)=a,g(x)=b , a,b \in \mathbb{Z}$

$ 2=ab \implies a= \pm1$ or $\pm2$

If $a= \pm1$ then $<2,x>=<1>$.Hence there exist $r(X) , s(X) \in \mathbb{Z}[X]$ such that $1= 2r(X) + Xs(X)$

If $a= \pm2$ then $<2,x>=<2>$.Hence there exist $v(X) \in \mathbb{Z}[X]$ such that $X= 2v(X)$

My confusion : why is $2= 2r(X) + Xs(X)$ not mentioned in the proof ?

**My thinking** : If $a= \pm2$ then $<2,x>=<2>$.Then there exist $r(X) , s(X) \in \mathbb{Z}[X]$ such that $2= 2r(X) + Xs(X)$

Now ,put $x=0$.Then $2=2r(0) + 0 s(0) \implies r(0)=1 \in \mathbb{Z}[x]$ i,e constant term will be $r(0)=1$

Tell me where I'm wrong?