Divide using synthetic division $$ ( 6x^{5}+55x^{4}+59x^{3}+31x^{2}+54x-15 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-8&6&55&59&31&54&-15\\& & -48& -56& -24& -56& \color{black}{16} \\ \hline &\color{blue}{6}&\color{blue}{7}&\color{blue}{3}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ 6x^{5}+55x^{4}+59x^{3}+31x^{2}+54x-15 }{ x+8 } = \color{blue}{6x^{4}+7x^{3}+3x^{2}+7x-2} ~+~ \frac{ \color{red}{ 1 } }{ x+8 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-8&\color{orangered}{ 6 }&55&59&31&54&-15\\& & & & & & \\ \hline &\color{orangered}{6}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & \color{blue}{-48} & & & & \\ \hline &\color{blue}{6}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 55 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-8&6&\color{orangered}{ 55 }&59&31&54&-15\\& & \color{orangered}{-48} & & & & \\ \hline &6&\color{orangered}{7}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 7 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & -48& \color{blue}{-56} & & & \\ \hline &6&\color{blue}{7}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 59 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrrr}-8&6&55&\color{orangered}{ 59 }&31&54&-15\\& & -48& \color{orangered}{-56} & & & \\ \hline &6&7&\color{orangered}{3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 3 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & -48& -56& \color{blue}{-24} & & \\ \hline &6&7&\color{blue}{3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-8&6&55&59&\color{orangered}{ 31 }&54&-15\\& & -48& -56& \color{orangered}{-24} & & \\ \hline &6&7&3&\color{orangered}{7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 7 } = \color{blue}{ -56 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & -48& -56& -24& \color{blue}{-56} & \\ \hline &6&7&3&\color{blue}{7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 54 } + \color{orangered}{ \left( -56 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-8&6&55&59&31&\color{orangered}{ 54 }&-15\\& & -48& -56& -24& \color{orangered}{-56} & \\ \hline &6&7&3&7&\color{orangered}{-2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-8}&6&55&59&31&54&-15\\& & -48& -56& -24& -56& \color{blue}{16} \\ \hline &6&7&3&7&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 16 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-8&6&55&59&31&54&\color{orangered}{ -15 }\\& & -48& -56& -24& -56& \color{orangered}{16} \\ \hline &\color{blue}{6}&\color{blue}{7}&\color{blue}{3}&\color{blue}{7}&\color{blue}{-2}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{4}+7x^{3}+3x^{2}+7x-2 } $ with a remainder of $ \color{red}{ 1 } $.