Answer
$\dfrac{-11-8\sqrt{2}}{14}$
Work Step by Step
Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $
\dfrac{3+\sqrt{8}}{2-2\sqrt{8}}
,$ is equivalent to
\begin{align*}\require{cancel}
&
\dfrac{3+\sqrt{8}}{2-2\sqrt{8}}\cdot\dfrac{2+2\sqrt{8}}{2+2\sqrt{8}}
\\\\&=
\dfrac{(3+\sqrt{8})(2+2\sqrt{8})}{2^2-(2\sqrt{8})^2}
&\left( \text{use }(a+b)(a-b)=a^2-b^2 \right)
\\\\&=
\dfrac{3(2)+3(2\sqrt{8})+\sqrt{8}(2)+\sqrt{8}(2\sqrt{8})}{2^2-(2\sqrt{8})^2}
&\left( \text{use FOIL} \right)
\\\\&=
\dfrac{6+6\sqrt{8}+2\sqrt{8}+2(8)}{4-4(8)}
\\\\&=
\dfrac{6+6\sqrt{8}+2\sqrt{8}+16}{4-32}
\\\\&=
\dfrac{22+8\sqrt{8}}{-28}
\\\\&=
\dfrac{22+8\sqrt{4\cdot2}}{-28}
\\\\&=
\dfrac{22+8\sqrt{(2)^2\cdot2}}{-28}
\\\\&=
\dfrac{22+8(2)\sqrt{2}}{-28}
\\\\&=
\dfrac{22+16\sqrt{2}}{-28}
\\\\&=
\dfrac{\cancel{22}^{-11}\cancel{+16}^{-8}\sqrt{2}}{\cancel{-28}^{14}}
&\left( \text{divide by }-2 \right)
\\\\&=
\dfrac{-11-8\sqrt{2}}{14}
.\end{align*}
Hence, the rationalized-denominator form of the given expression is $
\dfrac{-11-8\sqrt{2}}{14}
$.
