Divide using synthetic division $$ ( 6x^{3}+47x^{2}+2x+84 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-8&6&47&2&84\\& & -48& 8& \color{black}{-80} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+47x^{2}+2x+84 }{ x+8 } = \color{blue}{6x^{2}-x+10} ~+~ \frac{ \color{red}{ 4 } }{ x+8 } $$Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&47&2&84\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-8&\color{orangered}{ 6 }&47&2&84\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&47&2&84\\& & \color{blue}{-48} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 47 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-8&6&\color{orangered}{ 47 }&2&84\\& & \color{orangered}{-48} & & \\ \hline &6&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&47&2&84\\& & -48& \color{blue}{8} & \\ \hline &6&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 8 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-8&6&47&\color{orangered}{ 2 }&84\\& & -48& \color{orangered}{8} & \\ \hline &6&-1&\color{orangered}{10}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&47&2&84\\& & -48& 8& \color{blue}{-80} \\ \hline &6&-1&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 84 } + \color{orangered}{ \left( -80 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-8&6&47&2&\color{orangered}{ 84 }\\& & -48& 8& \color{orangered}{-80} \\ \hline &\color{blue}{6}&\color{blue}{-1}&\color{blue}{10}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-x+10 } $ with a remainder of $ \color{red}{ 4 } $.