Answer
$$\frac{1}{{3\left( {x - 4} \right)}} - \frac{1}{{3\left( {x + 2} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{2}{{{x^2} - 2x - 8}} \cr
& {\text{factoring}} \cr
& = \frac{2}{{\left( {x - 4} \right)\left( {x + 2} \right)}} \cr
& {\text{partial fraction decomposition}} \cr
& \frac{2}{{\left( {x - 4} \right)\left( {x + 2} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 2}} \cr
& 2 = A\left( {x + 2} \right) + B\left( {x - 4} \right) \cr
& {\text{letting }}x = 4 \cr
& 2 = A\left( {4 + 2} \right) + B\left( {4 - 4} \right) \cr
& 2 = A\left( 6 \right) \cr
& A = 1/3 \cr
& {\text{letting }}x = - 2 \cr
& 2 = A\left( { - 2 + 2} \right) + B\left( { - 2 - 4} \right) \cr
& 2 = B\left( { - 6} \right) \cr
& B = - 1/3 \cr
& {\text{substituting the values}} \cr
& \frac{A}{{x - 4}} + \frac{B}{{x + 2}} = \frac{{1/3}}{{x - 4}} - \frac{{1/3}}{{x + 2}} \cr
& = \frac{1}{{3\left( {x - 4} \right)}} - \frac{1}{{3\left( {x + 2} \right)}} \cr} $$
