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