How much information one root can give us about the other roots of a polynomial Suppose $a,b \in \mathbb{Q}$ and $f(X) \in \mathbb{Q}[X]$. In order to prove if $a+\sqrt{2}b$ is a root of $f(X)$, then $a-\sqrt{2}b$ is also a root, I did as follows.
I showed if $(a + \sqrt{2}b)^n=A_n+\sqrt{2}B_n$ then $(a - \sqrt{2}b)^n=A_n-\sqrt{2}B_n$ by induction on $n$, where $A_n, B_n \in \mathbb{Q}$. Then $f(a+\sqrt{2}b)=(c_0A_0+\cdots+c_nA_n)+\sqrt{2}(c_0B_0+\cdots+c_nB_n)$ implies $f(a-\sqrt{2}b)=(c_0A_0+\cdots+c_nA_n)-\sqrt{2}(c_0B_0+\cdots+c_nB_n)$, and the result is followed.
Now, my question is what can we say about roots of $f(X)$ if $a+\sqrt{\sqrt{2}}b$ is a root of $f(X)$?
Or, more generally, let $a,b \in R$, where $R$ is a domain, and $f(X) \in R[X]$. If $a+zb$ is a root of $f(X)$ where $z^i \not \in R$ for $1\le i<m-1$ and $z^m \in R$, then what can we say about roots of $f(X)$?

The minimal polynomial of $\sqrt{\sqrt{2}}$ is $X^4-2$.  If $f(a + \sqrt{\sqrt{2}} b) = 0$ where $f(X) \in \mathbb Q[X]$ and $b \ne 0$, that says $\sqrt{\sqrt{2}}$ is a root of
$f(a + X b)$, so $X^4 - 2 \mid f(a + X b)$, i.e. $\left(\frac{Y-a}{b}\right)^4 - 2 \mid f(Y)$.  Now $a - \sqrt{\sqrt{2}} b$ is a root of  $\left(\frac{Y-a}{b}\right)^4 - 2$, so that $f\left(a - \sqrt{\sqrt{2}} b\right) = 0$.
This still leaves the case $b = 0$, but that's trivial because $a + \sqrt{\sqrt{2}} b = a = a - \sqrt{\sqrt{2}} b$.
