Divide using synthetic division $$ ( 3x^{3}+5x^{2}-16x-12 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&5&-16&-12\\& & -6& 2& \color{black}{28} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-14}&\color{orangered}{16} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+5x^{2}-16x-12 }{ x+2 } = \color{blue}{3x^{2}-x-14} ~+~ \frac{ \color{red}{ 16 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&-16&-12\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&5&-16&-12\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&-16&-12\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 5 }&-16&-12\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&-16&-12\\& & -6& \color{blue}{2} & \\ \hline &3&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 2 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}-2&3&5&\color{orangered}{ -16 }&-12\\& & -6& \color{orangered}{2} & \\ \hline &3&-1&\color{orangered}{-14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 28 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&5&-16&-12\\& & -6& 2& \color{blue}{28} \\ \hline &3&-1&\color{blue}{-14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 28 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrrr}-2&3&5&-16&\color{orangered}{ -12 }\\& & -6& 2& \color{orangered}{28} \\ \hline &\color{blue}{3}&\color{blue}{-1}&\color{blue}{-14}&\color{orangered}{16} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-x-14 } $ with a remainder of $ \color{red}{ 16 } $.