The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&-1&-10&-8\\& & -2& 6& \color{black}{8} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x+2 $ is a factor of the $ x^{3}-x^{2}-10x-8 $.
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-1&-10&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&-1&-10&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-1&-10&-8\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ -1 }&-10&-8\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-1&-10&-8\\& & -2& \color{blue}{6} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 6 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-2&1&-1&\color{orangered}{ -10 }&-8\\& & -2& \color{orangered}{6} & \\ \hline &1&-3&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&-1&-10&-8\\& & -2& 6& \color{blue}{8} \\ \hline &1&-3&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 8 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}-2&1&-1&-10&\color{orangered}{ -8 }\\& & -2& 6& \color{orangered}{8} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.