Divide using synthetic division $$ ( 6x^{3}-38x^{2}-70x-92 ) \div ( x-8 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&6&-38&-70&-92\\& & 48& 80& \color{black}{80} \\ \hline &\color{blue}{6}&\color{blue}{10}&\color{blue}{10}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \dfrac{ 6x^{3}-38x^{2}-70x-92 }{ x-8 } = \color{blue}{6x^{2}+10x+10} \color{red}{~-~} \dfrac{ \color{red}{ 12 } }{ x-8 } $$Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&6&-38&-70&-92\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 6 }&-38&-70&-92\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 6 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&6&-38&-70&-92\\& & \color{blue}{48} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -38 } + \color{orangered}{ 48 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}8&6&\color{orangered}{ -38 }&-70&-92\\& & \color{orangered}{48} & & \\ \hline &6&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 10 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&6&-38&-70&-92\\& & 48& \color{blue}{80} & \\ \hline &6&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -70 } + \color{orangered}{ 80 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}8&6&-38&\color{orangered}{ -70 }&-92\\& & 48& \color{orangered}{80} & \\ \hline &6&10&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 10 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&6&-38&-70&-92\\& & 48& 80& \color{blue}{80} \\ \hline &6&10&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -92 } + \color{orangered}{ 80 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}8&6&-38&-70&\color{orangered}{ -92 }\\& & 48& 80& \color{orangered}{80} \\ \hline &\color{blue}{6}&\color{blue}{10}&\color{blue}{10}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+10x+10 } $ with a remainder of $ \color{red}{ -12 } $.