Check if $ x-4 $ is a factor of $ -x^{3}+8x+20 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&-1&0&8&20\\& & -4& -16& \color{black}{-32} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{-12} \end{array} $$Because the remainder $ \left( \color{red}{ -12 } \right) $ is not zero, we conclude that the $ x-4 $ is not a factor of $ -x^{3}+8x+20$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&0&8&20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ -1 }&0&8&20\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&0&8&20\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}4&-1&\color{orangered}{ 0 }&8&20\\& & \color{orangered}{-4} & & \\ \hline &-1&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&0&8&20\\& & -4& \color{blue}{-16} & \\ \hline &-1&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}4&-1&0&\color{orangered}{ 8 }&20\\& & -4& \color{orangered}{-16} & \\ \hline &-1&-4&\color{orangered}{-8}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&-1&0&8&20\\& & -4& -16& \color{blue}{-32} \\ \hline &-1&-4&\color{blue}{-8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}4&-1&0&8&\color{orangered}{ 20 }\\& & -4& -16& \color{orangered}{-32} \\ \hline &\color{blue}{-1}&\color{blue}{-4}&\color{blue}{-8}&\color{orangered}{-12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -12 }\right)$.