Divide using synthetic division $$ ( x^{3}-6x^{2}-4x+24 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&-6&-4&24\\& & 3& -9& \color{black}{-39} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-13}&\color{orangered}{-15} \end{array} $$The solution is:
$$ \frac{ x^{3}-6x^{2}-4x+24 }{ x-3 } = \color{blue}{x^{2}-3x-13} \color{red}{~-~} \frac{ \color{red}{ 15 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-6&-4&24\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&-6&-4&24\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-6&-4&24\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 3 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ -6 }&-4&24\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{-3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-6&-4&24\\& & 3& \color{blue}{-9} & \\ \hline &1&\color{blue}{-3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrr}3&1&-6&\color{orangered}{ -4 }&24\\& & 3& \color{orangered}{-9} & \\ \hline &1&-3&\color{orangered}{-13}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&-6&-4&24\\& & 3& -9& \color{blue}{-39} \\ \hline &1&-3&\color{blue}{-13}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -39 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrr}3&1&-6&-4&\color{orangered}{ 24 }\\& & 3& -9& \color{orangered}{-39} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-13}&\color{orangered}{-15} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-3x-13 } $ with a remainder of $ \color{red}{ -15 } $.