Check if $ x-4 $ is a factor of $ x^{3}-2x^{2}-16x-7 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&1&-2&-16&-7\\& & 4& 8& \color{black}{-32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-8}&\color{orangered}{-39} \end{array} $$Because the remainder $ \left( \color{red}{ -39 } \right) $ is not zero, we conclude that the $ x-4 $ is not a factor of $ x^{3}-2x^{2}-16x-7$.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-16&-7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 1 }&-2&-16&-7\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-16&-7\\& & \color{blue}{4} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 4 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}4&1&\color{orangered}{ -2 }&-16&-7\\& & \color{orangered}{4} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-16&-7\\& & 4& \color{blue}{8} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 8 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}4&1&-2&\color{orangered}{ -16 }&-7\\& & 4& \color{orangered}{8} & \\ \hline &1&2&\color{orangered}{-8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&1&-2&-16&-7\\& & 4& 8& \color{blue}{-32} \\ \hline &1&2&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -39 } $
$$ \begin{array}{c|rrrr}4&1&-2&-16&\color{orangered}{ -7 }\\& & 4& 8& \color{orangered}{-32} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{-8}&\color{orangered}{-39} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -39 }\right)$.