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