Divide using synthetic division $$ ( x^{4}+9x^{3}+14x^{2}+50x+9 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-8&1&9&14&50&9\\& & -8& -8& -48& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{6}&\color{blue}{2}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \frac{ x^{4}+9x^{3}+14x^{2}+50x+9 }{ x+8 } = \color{blue}{x^{3}+x^{2}+6x+2} \color{red}{~-~} \frac{ \color{red}{ 7 } }{ x+8 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&9&14&50&9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-8&\color{orangered}{ 1 }&9&14&50&9\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 1 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&9&14&50&9\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-8&1&\color{orangered}{ 9 }&14&50&9\\& & \color{orangered}{-8} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 1 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&9&14&50&9\\& & -8& \color{blue}{-8} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-8&1&9&\color{orangered}{ 14 }&50&9\\& & -8& \color{orangered}{-8} & & \\ \hline &1&1&\color{orangered}{6}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 6 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&9&14&50&9\\& & -8& -8& \color{blue}{-48} & \\ \hline &1&1&\color{blue}{6}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 50 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-8&1&9&14&\color{orangered}{ 50 }&9\\& & -8& -8& \color{orangered}{-48} & \\ \hline &1&1&6&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 2 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&9&14&50&9\\& & -8& -8& -48& \color{blue}{-16} \\ \hline &1&1&6&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-8&1&9&14&50&\color{orangered}{ 9 }\\& & -8& -8& -48& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{6}&\color{blue}{2}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+6x+2 } $ with a remainder of $ \color{red}{ -7 } $.