Check if $ x-2 $ is a factor of $ x^{5}-x^{4}-8x^{3}+11x^{2}+7x-10 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&-1&-8&11&7&-10\\& & 2& 2& -12& -2& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ x^{5}-x^{4}-8x^{3}+11x^{2}+7x-10 $.
Explanation
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}{ 2 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ -1 }&-8&11&7&-10\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-1&-8&11&7&-10\\& & 2& \color{blue}{2} & & & \\ \hline &1&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 2 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}2&1&-1&\color{orangered}{ -8 }&11&7&-10\\& & 2& \color{orangered}{2} & & & \\ \hline &1&1&\color{orangered}{-6}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-1&-8&11&7&-10\\& & 2& 2& \color{blue}{-12} & & \\ \hline &1&1&\color{blue}{-6}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrrr}2&1&-1&-8&\color{orangered}{ 11 }&7&-10\\& & 2& 2& \color{orangered}{-12} & & \\ \hline &1&1&-6&\color{orangered}{-1}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-1&-8&11&7&-10\\& & 2& 2& -12& \color{blue}{-2} & \\ \hline &1&1&-6&\color{blue}{-1}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}2&1&-1&-8&11&\color{orangered}{ 7 }&-10\\& & 2& 2& -12& \color{orangered}{-2} & \\ \hline &1&1&-6&-1&\color{orangered}{5}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&-1&-8&11&7&-10\\& & 2& 2& -12& -2& \color{blue}{10} \\ \hline &1&1&-6&-1&\color{blue}{5}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 10 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&1&-1&-8&11&7&\color{orangered}{ -10 }\\& & 2& 2& -12& -2& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.