Divide using synthetic division $$ ( x^{3}-15x^{2}+68x-96 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-15&68&-96\\& & 3& -36& \color{black}{96} \\ \hline &\color{blue}{1}&\color{blue}{-12}&\color{blue}{32}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{3}-15x^{2}+68x-96 }{ x-3 } = \color{blue}{x^{2}-12x+32} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-15&68&-96\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-15&68&-96\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-15&68&-96\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 3 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -15 }&68&-96\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-15&68&-96\\& & 3& \color{blue}{-36} & \\ \hline &1&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 68 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}3&1&-15&\color{orangered}{ 68 }&-96\\& & 3& \color{orangered}{-36} & \\ \hline &1&-12&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 32 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-15&68&-96\\& & 3& -36& \color{blue}{96} \\ \hline &1&-12&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -96 } + \color{orangered}{ 96 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&1&-15&68&\color{orangered}{ -96 }\\& & 3& -36& \color{orangered}{96} \\ \hline &\color{blue}{1}&\color{blue}{-12}&\color{blue}{32}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-12x+32 } $ with a remainder of $ \color{red}{ 0 } $.