Divide using synthetic division $$ ( 6x^{3}+42x^{2}-50x-20 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-8&6&42&-50&-20\\& & -48& 48& \color{black}{16} \\ \hline &\color{blue}{6}&\color{blue}{-6}&\color{blue}{-2}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+42x^{2}-50x-20 }{ x+8 } = \color{blue}{6x^{2}-6x-2} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+8 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&42&-50&-20\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-8&\color{orangered}{ 6 }&42&-50&-20\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&42&-50&-20\\& & \color{blue}{-48} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 42 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-8&6&\color{orangered}{ 42 }&-50&-20\\& & \color{orangered}{-48} & & \\ \hline &6&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&42&-50&-20\\& & -48& \color{blue}{48} & \\ \hline &6&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -50 } + \color{orangered}{ 48 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}-8&6&42&\color{orangered}{ -50 }&-20\\& & -48& \color{orangered}{48} & \\ \hline &6&-6&\color{orangered}{-2}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-8}&6&42&-50&-20\\& & -48& 48& \color{blue}{16} \\ \hline &6&-6&\color{blue}{-2}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 16 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}-8&6&42&-50&\color{orangered}{ -20 }\\& & -48& 48& \color{orangered}{16} \\ \hline &\color{blue}{6}&\color{blue}{-6}&\color{blue}{-2}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-6x-2 } $ with a remainder of $ \color{red}{ -4 } $.