Divide using synthetic division $$ ( x^{5}-23x^{3}+6x^{2}+112x-96 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}5&1&0&-23&6&112&-96\\& & 5& 25& 10& 80& \color{black}{960} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{2}&\color{blue}{16}&\color{blue}{192}&\color{orangered}{864} \end{array} $$The solution is:
$$ \frac{ x^{5}-23x^{3}+6x^{2}+112x-96 }{ x-5 } = \color{blue}{x^{4}+5x^{3}+2x^{2}+16x+192} ~+~ \frac{ \color{red}{ 864 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}5&\color{orangered}{ 1 }&0&-23&6&112&-96\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & \color{blue}{5} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrrr}5&1&\color{orangered}{ 0 }&-23&6&112&-96\\& & \color{orangered}{5} & & & & \\ \hline &1&\color{orangered}{5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & 5& \color{blue}{25} & & & \\ \hline &1&\color{blue}{5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -23 } + \color{orangered}{ 25 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}5&1&0&\color{orangered}{ -23 }&6&112&-96\\& & 5& \color{orangered}{25} & & & \\ \hline &1&5&\color{orangered}{2}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & 5& 25& \color{blue}{10} & & \\ \hline &1&5&\color{blue}{2}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 10 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrrrr}5&1&0&-23&\color{orangered}{ 6 }&112&-96\\& & 5& 25& \color{orangered}{10} & & \\ \hline &1&5&2&\color{orangered}{16}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 16 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & 5& 25& 10& \color{blue}{80} & \\ \hline &1&5&2&\color{blue}{16}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 112 } + \color{orangered}{ 80 } = \color{orangered}{ 192 } $
$$ \begin{array}{c|rrrrrr}5&1&0&-23&6&\color{orangered}{ 112 }&-96\\& & 5& 25& 10& \color{orangered}{80} & \\ \hline &1&5&2&16&\color{orangered}{192}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 192 } = \color{blue}{ 960 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{5}&1&0&-23&6&112&-96\\& & 5& 25& 10& 80& \color{blue}{960} \\ \hline &1&5&2&16&\color{blue}{192}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -96 } + \color{orangered}{ 960 } = \color{orangered}{ 864 } $
$$ \begin{array}{c|rrrrrr}5&1&0&-23&6&112&\color{orangered}{ -96 }\\& & 5& 25& 10& 80& \color{orangered}{960} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{2}&\color{blue}{16}&\color{blue}{192}&\color{orangered}{864} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+5x^{3}+2x^{2}+16x+192 } $ with a remainder of $ \color{red}{ 864 } $.