Divide using synthetic division $$ ( 12x^{3}+5x^{2}-11x-6 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&12&5&-11&-6\\& & -24& 38& \color{black}{-54} \\ \hline &\color{blue}{12}&\color{blue}{-19}&\color{blue}{27}&\color{orangered}{-60} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+5x^{2}-11x-6 }{ x+2 } = \color{blue}{12x^{2}-19x+27} \color{red}{~-~} \frac{ \color{red}{ 60 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&5&-11&-6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 12 }&5&-11&-6\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&5&-11&-6\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-2&12&\color{orangered}{ 5 }&-11&-6\\& & \color{orangered}{-24} & & \\ \hline &12&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&5&-11&-6\\& & -24& \color{blue}{38} & \\ \hline &12&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 38 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrr}-2&12&5&\color{orangered}{ -11 }&-6\\& & -24& \color{orangered}{38} & \\ \hline &12&-19&\color{orangered}{27}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 27 } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&5&-11&-6\\& & -24& 38& \color{blue}{-54} \\ \hline &12&-19&\color{blue}{27}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -60 } $
$$ \begin{array}{c|rrrr}-2&12&5&-11&\color{orangered}{ -6 }\\& & -24& 38& \color{orangered}{-54} \\ \hline &\color{blue}{12}&\color{blue}{-19}&\color{blue}{27}&\color{orangered}{-60} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}-19x+27 } $ with a remainder of $ \color{red}{ -60 } $.