HMMT February

$$\begin{aligned} x^{4}-4 x^{3}-4 x^{2}+16 x-8 & =\left(x^{4}-4 x^{3}+4 x^{2}\right)-\left(8 x^{2}-16 x+8\right) \\ & =x^{2}(x-2)^{2}-8(x-1)^{2} \\ & =\left(x^{2}-2 x\right)^{2}-(2 \sqrt{2} x-2 \sqrt{2})^{2} \\ & =\left(x^{2}-(2+2 \sqrt{2}) x+2 \sqrt{2}\right)\left(x^{2}-(2-2 \sqrt{2}) x-2 \sqrt{2}\right) \end{aligned}$$ But noting that $(1+\sqrt{2})^{2}=3+2 \sqrt{2}$ and completing the square, $$\begin{aligned} x^{2}-(2+2 \sqrt{2}) x+2 \sqrt{2} & =x^{2}-(2+2 \sqrt{2}) x+3+2 \sqrt{2}-3 \\ & =(x-(1+\sqrt{2}))^{2}-(\sqrt{3})^{2} \\ & =(x-1-\sqrt{2}+\sqrt{3})(x-1-\sqrt{2}-\sqrt{3}) \end{aligned}$$ Likewise, $$x^{2}-(2-2 \sqrt{2}) x-2 \sqrt{2}=(x-1+\sqrt{2}+\sqrt{3})(x-1+\sqrt{2}-\sqrt{3})$$ so the roots of the quartic are $1 \pm \sqrt{2} \pm \sqrt{3}$. Only one of these is negative, namely $1-\sqrt{2}-\sqrt{3}$, so the sum of the absolute values of the roots is $$(1+\sqrt{2}+\sqrt{3})+(1+\sqrt{2}-\sqrt{3})+(1-\sqrt{2}+\sqrt{3})-(1-\sqrt{2}-\sqrt{3})=2+2 \sqrt{2}+2 \sqrt{3}$$