Divide using synthetic division $$ ( x^{4}+9x^{3}+14x^{2}-8x-16 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&9&14&-8&-16\\& & -2& -14& 0& \color{black}{16} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}+9x^{3}+14x^{2}-8x-16 }{ x+2 } = \color{blue}{x^{3}+7x^{2}-8} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&9&14&-8&-16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&9&14&-8&-16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&9&14&-8&-16\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 9 }&14&-8&-16\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{7}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 7 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&9&14&-8&-16\\& & -2& \color{blue}{-14} & & \\ \hline &1&\color{blue}{7}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&9&\color{orangered}{ 14 }&-8&-16\\& & -2& \color{orangered}{-14} & & \\ \hline &1&7&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&9&14&-8&-16\\& & -2& -14& \color{blue}{0} & \\ \hline &1&7&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 0 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}-2&1&9&14&\color{orangered}{ -8 }&-16\\& & -2& -14& \color{orangered}{0} & \\ \hline &1&7&0&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&9&14&-8&-16\\& & -2& -14& 0& \color{blue}{16} \\ \hline &1&7&0&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 16 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&1&9&14&-8&\color{orangered}{ -16 }\\& & -2& -14& 0& \color{orangered}{16} \\ \hline &\color{blue}{1}&\color{blue}{7}&\color{blue}{0}&\color{blue}{-8}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+7x^{2}-8 } $ with a remainder of $ \color{red}{ 0 } $.