Divide using synthetic division $$ ( 3x^{5}+6x^{2}+7 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-2&3&0&0&6&0&7\\& & -6& 12& -24& 36& \color{black}{-72} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-18}&\color{blue}{36}&\color{orangered}{-65} \end{array} $$The solution is:
$$ \frac{ 3x^{5}+6x^{2}+7 }{ x+2 } = \color{blue}{3x^{4}-6x^{3}+12x^{2}-18x+36} \color{red}{~-~} \frac{ \color{red}{ 65 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-2&\color{orangered}{ 3 }&0&0&6&0&7\\& & & & & & \\ \hline &\color{orangered}{3}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & \color{blue}{-6} & & & & \\ \hline &\color{blue}{3}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrrr}-2&3&\color{orangered}{ 0 }&0&6&0&7\\& & \color{orangered}{-6} & & & & \\ \hline &3&\color{orangered}{-6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & -6& \color{blue}{12} & & & \\ \hline &3&\color{blue}{-6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}-2&3&0&\color{orangered}{ 0 }&6&0&7\\& & -6& \color{orangered}{12} & & & \\ \hline &3&-6&\color{orangered}{12}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & -6& 12& \color{blue}{-24} & & \\ \hline &3&-6&\color{blue}{12}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrrr}-2&3&0&0&\color{orangered}{ 6 }&0&7\\& & -6& 12& \color{orangered}{-24} & & \\ \hline &3&-6&12&\color{orangered}{-18}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & -6& 12& -24& \color{blue}{36} & \\ \hline &3&-6&12&\color{blue}{-18}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 36 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrrr}-2&3&0&0&6&\color{orangered}{ 0 }&7\\& & -6& 12& -24& \color{orangered}{36} & \\ \hline &3&-6&12&-18&\color{orangered}{36}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 36 } = \color{blue}{ -72 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-2}&3&0&0&6&0&7\\& & -6& 12& -24& 36& \color{blue}{-72} \\ \hline &3&-6&12&-18&\color{blue}{36}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -72 \right) } = \color{orangered}{ -65 } $
$$ \begin{array}{c|rrrrrr}-2&3&0&0&6&0&\color{orangered}{ 7 }\\& & -6& 12& -24& 36& \color{orangered}{-72} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{12}&\color{blue}{-18}&\color{blue}{36}&\color{orangered}{-65} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{4}-6x^{3}+12x^{2}-18x+36 } $ with a remainder of $ \color{red}{ -65 } $.