Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$. 
Let $f(x)$ be a polynomial in $\mathbb{Q}[x]$. Prove that if $a + \sqrt2 b$ is a zero of $f(x)$ then so is $a - \sqrt2b$.

Can I treat $a + \sqrt2 b$ like a complex number? Then I can do something like
let $f(x) = \sum a_ix^i$, and  $z = a+\sqrt2b$
Then $f(\bar z) = \sum a_i \bar z^i = ... = \bar{f(z)}$.
If not.. what else can I try?
 A: Hint: employ conjugation  as here or, if that's unknown, give a direct proof, e.g. as below.
$a\pm\sqrt2\,b\,$ is a root of $f(x)\!\iff\! \pm\sqrt 2\,$ is a root of $\,f(a+bx)=: \bar f(x)\in\Bbb Q[x]$
Thus, equivalently, we seek  $\,\bar f(\sqrt 2) = 0\,\Rightarrow\, \bar f(-\sqrt 2) = 0$.
Now split $\bar f(x) = g(x^2) + h(x^2)\,x\,\ $ into even + odd parts $ $
(cf. here and here)
so $\ 0 = \bar f(\sqrt 2) = g(2)\ +\ h(2) \sqrt 2\,\Rightarrow\, \color{#c00}{h(2)\!=\!0}\Rightarrow \color{#c00}{g(2)\!=\!0},\,$ by $\,g(2),h(2)\in\Bbb Q\not\ni \sqrt 2$
thus $\,\ \bar f(-\sqrt 2) =  \color{#c00}{g(2)\ -\ h(2)} \sqrt 2 = 0. \ \ \ \rm\small QED$
A: We can divide $f$ by $\left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right) = x^2-2ax+a^2-2b^2 \in \Bbb{Q}[x]$ to obtain
$$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x) + cx+d$$
for some $q \in \Bbb{Q}[x]$ and $c,d \in \Bbb{Q}$. Plugging in $x = a+\sqrt{2}b$ gives
$$0 = f(a+\sqrt{2}b) = c(a+\sqrt{2}b) + d = (ca+d) + \sqrt{2}cb$$
which by irrationality of $\sqrt{2}$ implies $ca+d = cb = 0$. Assuming $b \ne 0$, this implies $c = 0$ and then $d = 0$. Therefore $$f(x) = \left(x-(a+\sqrt{2}b)\right)\left(x-(a-\sqrt{2}b)\right)q(x)$$
so in particular $f(a-\sqrt{2}b) = 0$.
