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