Divide using synthetic division $$ ( x^{3}-4x^{2}-5x-42 ) \div ( x-6 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}6&1&-4&-5&-42\\& & 6& 12& \color{black}{42} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{0} \end{array} $$The solution is:
$$ \dfrac{ x^{3}-4x^{2}-5x-42 }{ x-6 } = \color{blue}{x^{2}+2x+7} $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -6 = 0 $ ( $ x = \color{blue}{ 6 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-4&-5&-42\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}6&\color{orangered}{ 1 }&-4&-5&-42\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 1 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-4&-5&-42\\& & \color{blue}{6} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 6 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}6&1&\color{orangered}{ -4 }&-5&-42\\& & \color{orangered}{6} & & \\ \hline &1&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 2 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-4&-5&-42\\& & 6& \color{blue}{12} & \\ \hline &1&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 12 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}6&1&-4&\color{orangered}{ -5 }&-42\\& & 6& \color{orangered}{12} & \\ \hline &1&2&\color{orangered}{7}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 6 } \cdot \color{blue}{ 7 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{6}&1&-4&-5&-42\\& & 6& 12& \color{blue}{42} \\ \hline &1&2&\color{blue}{7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ 42 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}6&1&-4&-5&\color{orangered}{ -42 }\\& & 6& 12& \color{orangered}{42} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{7}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+2x+7 } $ with a remainder of $ \color{red}{ 0 } $.