The synthetic division table is:
$$ \begin{array}{c|rrr}-1&6&-2&2\\& & -6& \color{black}{8} \\ \hline &\color{blue}{6}&\color{blue}{-8}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ 6x^{2}-2x+2 }{ x+1 } = \color{blue}{6x-8} ~+~ \frac{ \color{red}{ 10 } }{ x+1 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-2&2\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}-1&\color{orangered}{ 6 }&-2&2\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 6 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-2&2\\& & \color{blue}{-6} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrr}-1&6&\color{orangered}{ -2 }&2\\& & \color{orangered}{-6} & \\ \hline &6&\color{orangered}{-8}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrr}\color{blue}{-1}&6&-2&2\\& & -6& \color{blue}{8} \\ \hline &6&\color{blue}{-8}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 8 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrr}-1&6&-2&\color{orangered}{ 2 }\\& & -6& \color{orangered}{8} \\ \hline &\color{blue}{6}&\color{blue}{-8}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x-8 } $ with a remainder of $ \color{red}{ 10 } $.