Divide using synthetic division $$ ( x^{5}-47x^{3}-16x^{2}+8x+52 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}7&1&0&-47&-16&8&52\\& & 7& 49& 14& -14& \color{black}{-42} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{-6}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{5}-47x^{3}-16x^{2}+8x+52 }{ x-7 } = \color{blue}{x^{4}+7x^{3}+2x^{2}-2x-6} ~+~ \frac{ \color{red}{ 10 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}7&\color{orangered}{ 1 }&0&-47&-16&8&52\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & \color{blue}{7} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 7 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}7&1&\color{orangered}{ 0 }&-47&-16&8&52\\& & \color{orangered}{7} & & & & \\ \hline &1&\color{orangered}{7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 7 } = \color{blue}{ 49 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & 7& \color{blue}{49} & & & \\ \hline &1&\color{blue}{7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -47 } + \color{orangered}{ 49 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}7&1&0&\color{orangered}{ -47 }&-16&8&52\\& & 7& \color{orangered}{49} & & & \\ \hline &1&7&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 2 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & 7& 49& \color{blue}{14} & & \\ \hline &1&7&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 14 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}7&1&0&-47&\color{orangered}{ -16 }&8&52\\& & 7& 49& \color{orangered}{14} & & \\ \hline &1&7&2&\color{orangered}{-2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & 7& 49& 14& \color{blue}{-14} & \\ \hline &1&7&2&\color{blue}{-2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}7&1&0&-47&-16&\color{orangered}{ 8 }&52\\& & 7& 49& 14& \color{orangered}{-14} & \\ \hline &1&7&2&-2&\color{orangered}{-6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{7}&1&0&-47&-16&8&52\\& & 7& 49& 14& -14& \color{blue}{-42} \\ \hline &1&7&2&-2&\color{blue}{-6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 52 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}7&1&0&-47&-16&8&\color{orangered}{ 52 }\\& & 7& 49& 14& -14& \color{orangered}{-42} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{2}&\color{blue}{-2}&\color{blue}{-6}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+7x^{3}+2x^{2}-2x-6 } $ with a remainder of $ \color{red}{ 10 } $.