<s>[INST] Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-8r^2+8=0.\tag{$\star$}$$ 
The rational zeros theorem tells us that the only possible rational solutions to ($\star$) are $±1$, $±2$, $±4$, $±8$.
I do not know where to go from here. Please help me complete this proof.   
 [/INST] Check your second line, $r^2+2r\sqrt{2} + 2 = 6 + 4\sqrt{2} \ne r^2-4 -2\sqrt{2}$. 
It does, however, equal $ (r-2) (r+2 \sqrt{2}+2) = 0$ which would imply that $r = 2$ or $r= -2(1+\sqrt{2})$.
Thus $2$ is the only rational root of your polynomial and $\sqrt{6+4\sqrt{2}} -\sqrt{2}= 2 $ (since $r =\sqrt{6+4\sqrt{2}} -\sqrt{2}$ and $(r-2)=\sqrt{6+4\sqrt{2}} -\sqrt{2} -2 = 0$). 
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