The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&1&0&-14&-8\\& & -1& 1& \color{black}{13} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-13}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ x^{3}-14x-8 }{ x+1 } = \color{blue}{x^{2}-x-13} ~+~ \frac{ \color{red}{ 5 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&0&-14&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 1 }&0&-14&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&0&-14&-8\\& & \color{blue}{-1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-1&1&\color{orangered}{ 0 }&-14&-8\\& & \color{orangered}{-1} & & \\ \hline &1&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&0&-14&-8\\& & -1& \color{blue}{1} & \\ \hline &1&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 1 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}-1&1&0&\color{orangered}{ -14 }&-8\\& & -1& \color{orangered}{1} & \\ \hline &1&-1&\color{orangered}{-13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ 13 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&1&0&-14&-8\\& & -1& 1& \color{blue}{13} \\ \hline &1&-1&\color{blue}{-13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 13 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-1&1&0&-14&\color{orangered}{ -8 }\\& & -1& 1& \color{orangered}{13} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-13}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-x-13 } $ with a remainder of $ \color{red}{ 5 } $.