Divide using synthetic division $$ ( 6x^{3}+x^{2}+4x+4 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&6&1&4&4\\& & -12& 22& \color{black}{-52} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{26}&\color{orangered}{-48} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+x^{2}+4x+4 }{ x+2 } = \color{blue}{6x^{2}-11x+26} \color{red}{~-~} \frac{ \color{red}{ 48 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&1&4&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 6 }&1&4&4\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 6 } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&1&4&4\\& & \color{blue}{-12} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -11 } $
$$ \begin{array}{c|rrrr}-2&6&\color{orangered}{ 1 }&4&4\\& & \color{orangered}{-12} & & \\ \hline &6&\color{orangered}{-11}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -11 \right) } = \color{blue}{ 22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&1&4&4\\& & -12& \color{blue}{22} & \\ \hline &6&\color{blue}{-11}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 22 } = \color{orangered}{ 26 } $
$$ \begin{array}{c|rrrr}-2&6&1&\color{orangered}{ 4 }&4\\& & -12& \color{orangered}{22} & \\ \hline &6&-11&\color{orangered}{26}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 26 } = \color{blue}{ -52 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&6&1&4&4\\& & -12& 22& \color{blue}{-52} \\ \hline &6&-11&\color{blue}{26}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -52 \right) } = \color{orangered}{ -48 } $
$$ \begin{array}{c|rrrr}-2&6&1&4&\color{orangered}{ 4 }\\& & -12& 22& \color{orangered}{-52} \\ \hline &\color{blue}{6}&\color{blue}{-11}&\color{blue}{26}&\color{orangered}{-48} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}-11x+26 } $ with a remainder of $ \color{red}{ -48 } $.