Check if $ x+2 $ is a factor of $ x^{5}-x^{4}-8x^{3}+11x^{2}+7x-10 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&1&-1&-8&11&7&-10\\& & -2& 6& 4& -30& \color{black}{46} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-2}&\color{blue}{15}&\color{blue}{-23}&\color{orangered}{36} \end{array} $$Because the remainder $ \left( \color{red}{ 36 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ x^{5}-x^{4}-8x^{3}+11x^{2}+7x-10$.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 1 }&-1&-8&11&7&-10\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & \color{blue}{-2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-2&1&\color{orangered}{ -1 }&-8&11&7&-10\\& & \color{orangered}{-2} & & & & \\ \hline &1&\color{orangered}{-3}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & -2& \color{blue}{6} & & & \\ \hline &1&\color{blue}{-3}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 6 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-2&1&-1&\color{orangered}{ -8 }&11&7&-10\\& & -2& \color{orangered}{6} & & & \\ \hline &1&-3&\color{orangered}{-2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & -2& 6& \color{blue}{4} & & \\ \hline &1&-3&\color{blue}{-2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 4 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrrr}-2&1&-1&-8&\color{orangered}{ 11 }&7&-10\\& & -2& 6& \color{orangered}{4} & & \\ \hline &1&-3&-2&\color{orangered}{15}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 15 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & -2& 6& 4& \color{blue}{-30} & \\ \hline &1&-3&-2&\color{blue}{15}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ -23 } $
$$ \begin{array}{c|rrrrrr}-2&1&-1&-8&11&\color{orangered}{ 7 }&-10\\& & -2& 6& 4& \color{orangered}{-30} & \\ \hline &1&-3&-2&15&\color{orangered}{-23}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -23 \right) } = \color{blue}{ 46 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&1&-1&-8&11&7&-10\\& & -2& 6& 4& -30& \color{blue}{46} \\ \hline &1&-3&-2&15&\color{blue}{-23}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 46 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrrr}-2&1&-1&-8&11&7&\color{orangered}{ -10 }\\& & -2& 6& 4& -30& \color{orangered}{46} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-2}&\color{blue}{15}&\color{blue}{-23}&\color{orangered}{36} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 36 }\right)$.