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