Divide using synthetic division $$ ( x^{3}+12x^{2}+20x-96 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&12&20&-96\\& & 2& 28& \color{black}{96} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{48}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}+12x^{2}+20x-96 }{ x-2 } = \color{blue}{x^{2}+14x+48} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&12&20&-96\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&12&20&-96\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&12&20&-96\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 2 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 12 }&20&-96\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{14}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 14 } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&12&20&-96\\& & 2& \color{blue}{28} & \\ \hline &1&\color{blue}{14}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ 28 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrr}2&1&12&\color{orangered}{ 20 }&-96\\& & 2& \color{orangered}{28} & \\ \hline &1&14&\color{orangered}{48}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 48 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&12&20&-96\\& & 2& 28& \color{blue}{96} \\ \hline &1&14&\color{blue}{48}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -96 } + \color{orangered}{ 96 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}2&1&12&20&\color{orangered}{ -96 }\\& & 2& 28& \color{orangered}{96} \\ \hline &\color{blue}{1}&\color{blue}{14}&\color{blue}{48}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+14x+48 } $ with a remainder of $ \color{red}{ 0 } $.