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