Check if $ x-2 $ is a factor of $ x^{3}-20x-8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&0&-20&-8\\& & 2& 4& \color{black}{-32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-16}&\color{orangered}{-40} \end{array} $$Because the remainder $ \left( \color{red}{ -40 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ x^{3}-20x-8$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&0&-20&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&0&-20&-8\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&0&-20&-8\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 0 }&-20&-8\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&0&-20&-8\\& & 2& \color{blue}{4} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 4 } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrr}2&1&0&\color{orangered}{ -20 }&-8\\& & 2& \color{orangered}{4} & \\ \hline &1&2&\color{orangered}{-16}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&0&-20&-8\\& & 2& 4& \color{blue}{-32} \\ \hline &1&2&\color{blue}{-16}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrr}2&1&0&-20&\color{orangered}{ -8 }\\& & 2& 4& \color{orangered}{-32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-16}&\color{orangered}{-40} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -40 }\right)$.