Find the least degree polynomial equation with rational coefficients whose roots is given $\sqrt2-1$
My book has given the solution as
$x=(\sqrt2-1)$
$(x+1)^2 = 2$
I don't understand how this became the required polynomial equation.......I can't even determine if this has $(\sqrt2-1)$ even as a solution by looking at it. They have done the similar for obtaining $1+\sqrt2+\sqrt3$ as a root. For another polynomial least degree
Let’s do that step by step. Write
$$x=\sqrt{2}-1$$
This is equivalent to
$$x+1=\sqrt{2}$$
This means that $x$ is one of the two real numbers (the other one is $-1-\sqrt{2}$) such that
$$(x+1)^2=2$$
This rewrites as
$$x^2+2x-1=0$$
Is this the smallest degree with rational coefficients ? Yes because over the real numbers the smallest degree possible equation would be
$$x-\sqrt{2}+1=0$$
The coefficients cannot be rational because this would mean $\sqrt{2}$ is rational