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