Divide using synthetic division $$ ( x^{3}+8x^{2}+16x+14 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&8&16&14\\& & -2& -12& \color{black}{-8} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{4}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ x^{3}+8x^{2}+16x+14 }{ x+2 } = \color{blue}{x^{2}+6x+4} ~+~ \frac{ \color{red}{ 6 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&8&16&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&8&16&14\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&8&16&14\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 8 }&16&14\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&8&16&14\\& & -2& \color{blue}{-12} & \\ \hline &1&\color{blue}{6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-2&1&8&\color{orangered}{ 16 }&14\\& & -2& \color{orangered}{-12} & \\ \hline &1&6&\color{orangered}{4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&8&16&14\\& & -2& -12& \color{blue}{-8} \\ \hline &1&6&\color{blue}{4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrr}-2&1&8&16&\color{orangered}{ 14 }\\& & -2& -12& \color{orangered}{-8} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{4}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+6x+4 } $ with a remainder of $ \color{red}{ 6 } $.