Check if $ x-2 $ is a factor of $ x^{5}+7x^{4}+10x^{3}+6x^{2}+4x-16 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&1&7&10&6&4&-16\\& & 2& 18& 56& 124& \color{black}{256} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{28}&\color{blue}{62}&\color{blue}{128}&\color{orangered}{240} \end{array} $$Because the remainder $ \left( \color{red}{ 240 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ x^{5}+7x^{4}+10x^{3}+6x^{2}+4x-16$.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 1 }&7&10&6&4&-16\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & \color{blue}{2} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrrr}2&1&\color{orangered}{ 7 }&10&6&4&-16\\& & \color{orangered}{2} & & & & \\ \hline &1&\color{orangered}{9}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & 2& \color{blue}{18} & & & \\ \hline &1&\color{blue}{9}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 18 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrrrr}2&1&7&\color{orangered}{ 10 }&6&4&-16\\& & 2& \color{orangered}{18} & & & \\ \hline &1&9&\color{orangered}{28}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 28 } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & 2& 18& \color{blue}{56} & & \\ \hline &1&9&\color{blue}{28}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 56 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrrrr}2&1&7&10&\color{orangered}{ 6 }&4&-16\\& & 2& 18& \color{orangered}{56} & & \\ \hline &1&9&28&\color{orangered}{62}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 62 } = \color{blue}{ 124 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & 2& 18& 56& \color{blue}{124} & \\ \hline &1&9&28&\color{blue}{62}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 124 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrrr}2&1&7&10&6&\color{orangered}{ 4 }&-16\\& & 2& 18& 56& \color{orangered}{124} & \\ \hline &1&9&28&62&\color{orangered}{128}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 128 } = \color{blue}{ 256 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&1&7&10&6&4&-16\\& & 2& 18& 56& 124& \color{blue}{256} \\ \hline &1&9&28&62&\color{blue}{128}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 256 } = \color{orangered}{ 240 } $
$$ \begin{array}{c|rrrrrr}2&1&7&10&6&4&\color{orangered}{ -16 }\\& & 2& 18& 56& 124& \color{orangered}{256} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{28}&\color{blue}{62}&\color{blue}{128}&\color{orangered}{240} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 240 }\right)$.