Divide using synthetic division $$ ( x^{5}-4x^{4}-22x^{3}+87x-48 ) \div ( x-9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}9&1&-4&-22&0&87&-48\\& & 9& 45& 207& 1863& \color{black}{17550} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{23}&\color{blue}{207}&\color{blue}{1950}&\color{orangered}{17502} \end{array} $$The solution is:
$$ \frac{ x^{5}-4x^{4}-22x^{3}+87x-48 }{ x-9 } = \color{blue}{x^{4}+5x^{3}+23x^{2}+207x+1950} ~+~ \frac{ \color{red}{ 17502 } }{ x-9 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -9 = 0 $ ( $ x = \color{blue}{ 9 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}9&\color{orangered}{ 1 }&-4&-22&0&87&-48\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 1 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & \color{blue}{9} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}9&1&\color{orangered}{ -4 }&-22&0&87&-48\\& & \color{orangered}{9} & & & & \\ \hline &1&\color{orangered}{5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 5 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & 9& \color{blue}{45} & & & \\ \hline &1&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 45 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrrrr}9&1&-4&\color{orangered}{ -22 }&0&87&-48\\& & 9& \color{orangered}{45} & & & \\ \hline &1&5&\color{orangered}{23}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 23 } = \color{blue}{ 207 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & 9& 45& \color{blue}{207} & & \\ \hline &1&5&\color{blue}{23}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 207 } = \color{orangered}{ 207 } $
$$ \begin{array}{c|rrrrrr}9&1&-4&-22&\color{orangered}{ 0 }&87&-48\\& & 9& 45& \color{orangered}{207} & & \\ \hline &1&5&23&\color{orangered}{207}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 207 } = \color{blue}{ 1863 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & 9& 45& 207& \color{blue}{1863} & \\ \hline &1&5&23&\color{blue}{207}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 87 } + \color{orangered}{ 1863 } = \color{orangered}{ 1950 } $
$$ \begin{array}{c|rrrrrr}9&1&-4&-22&0&\color{orangered}{ 87 }&-48\\& & 9& 45& 207& \color{orangered}{1863} & \\ \hline &1&5&23&207&\color{orangered}{1950}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 9 } \cdot \color{blue}{ 1950 } = \color{blue}{ 17550 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{9}&1&-4&-22&0&87&-48\\& & 9& 45& 207& 1863& \color{blue}{17550} \\ \hline &1&5&23&207&\color{blue}{1950}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -48 } + \color{orangered}{ 17550 } = \color{orangered}{ 17502 } $
$$ \begin{array}{c|rrrrrr}9&1&-4&-22&0&87&\color{orangered}{ -48 }\\& & 9& 45& 207& 1863& \color{orangered}{17550} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{23}&\color{blue}{207}&\color{blue}{1950}&\color{orangered}{17502} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+5x^{3}+23x^{2}+207x+1950 } $ with a remainder of $ \color{red}{ 17502 } $.