Check if $ x+1 $ is a factor of $ x^{4}+x^{3}-6x^{2}-x+5 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}-1&1&1&-6&-1&5\\& & -1& 0& 6& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-6}&\color{blue}{5}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x+1 $ is a factor of the $ x^{4}+x^{3}-6x^{2}-x+5 $.

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.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&1&-6&-1&5\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&1&-6&-1&5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&1&-6&-1&5\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ 1 }&-6&-1&5\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&1&-6&-1&5\\& & -1& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $

$$ \begin{array}{c|rrrrr}-1&1&1&\color{orangered}{ -6 }&-1&5\\& & -1& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{-6}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 6 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&1&-6&-1&5\\& & -1& 0& \color{blue}{6} & \\ \hline &1&0&\color{blue}{-6}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 6 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrrr}-1&1&1&-6&\color{orangered}{ -1 }&5\\& & -1& 0& \color{orangered}{6} & \\ \hline &1&0&-6&\color{orangered}{5}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 5 } = \color{blue}{ -5 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&1&-6&-1&5\\& & -1& 0& 6& \color{blue}{-5} \\ \hline &1&0&-6&\color{blue}{5}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}-1&1&1&-6&-1&\color{orangered}{ 5 }\\& & -1& 0& 6& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{-6}&\color{blue}{5}&\color{orangered}{0} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.

Synthetic Division Calculator