Divide using synthetic division $$ ( x^{4}+10x^{3}+12x^{2}+24x ) \div ( x+9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-9&1&10&12&24&0\\& & -9& -9& -27& \color{black}{27} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-3}&\color{orangered}{27} \end{array} $$The solution is:
$$ \frac{ x^{4}+10x^{3}+12x^{2}+24x }{ x+9 } = \color{blue}{x^{3}+x^{2}+3x-3} ~+~ \frac{ \color{red}{ 27 } }{ x+9 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 9 = 0 $ ( $ x = \color{blue}{ -9 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-9}&1&10&12&24&0\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-9&\color{orangered}{ 1 }&10&12&24&0\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 1 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-9}&1&10&12&24&0\\& & \color{blue}{-9} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-9&1&\color{orangered}{ 10 }&12&24&0\\& & \color{orangered}{-9} & & & \\ \hline &1&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 1 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-9}&1&10&12&24&0\\& & -9& \color{blue}{-9} & & \\ \hline &1&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-9&1&10&\color{orangered}{ 12 }&24&0\\& & -9& \color{orangered}{-9} & & \\ \hline &1&1&\color{orangered}{3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 3 } = \color{blue}{ -27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-9}&1&10&12&24&0\\& & -9& -9& \color{blue}{-27} & \\ \hline &1&1&\color{blue}{3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 24 } + \color{orangered}{ \left( -27 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-9&1&10&12&\color{orangered}{ 24 }&0\\& & -9& -9& \color{orangered}{-27} & \\ \hline &1&1&3&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-9}&1&10&12&24&0\\& & -9& -9& -27& \color{blue}{27} \\ \hline &1&1&3&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 27 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}-9&1&10&12&24&\color{orangered}{ 0 }\\& & -9& -9& -27& \color{orangered}{27} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{3}&\color{blue}{-3}&\color{orangered}{27} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+x^{2}+3x-3 } $ with a remainder of $ \color{red}{ 27 } $.