We are asked to determine whether the equation $f(x)=8x^2+ x+ 1$ has a root in $\mathbb{Q}_2$. Now, I immediately think to apply Hensel's Lemma, with $\alpha=1$, by which I mean $f(\alpha)\equiv 0 \pmod{2}$, and $f'(\alpha)\equiv 1 \pmod{2}$, thus we can find $\alpha'\in \mathbb{Q}_2$ with $f(\alpha')=0\in \mathbb{Q}_2$.
This seemed fine, but in the given solution, the lecturer states that it is necessary to make the transformation $y=8x$, and proceed from there, letting $g=8f$, and then $g(y)=y^2+y+8$, and using $\alpha =1$.
I can only presume that there is a special property of $\mathbb{Q}_2$ that I am missing.
Explicitly, my question is why do we have to make this transformation; is it not sufficient to use my first method?
There’s another method, which I like because it’s well-suited to computation. From the relation $\alpha=-1-8\alpha^2$ that’s satisfied by a root of your polynomial, you derive a recursive method of finding $\alpha$, using the Banach Fixed Point Theorem. Start with $a_0=0$, and form $a_i=-1-8a_{i-1}^2$. Gives you your unit root, by taking $\lim_ia_i$.