Divide using synthetic division $$ ( 12x^{3}+26x^{2}-15x-18 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&12&26&-15&-18\\& & -24& -4& \color{black}{38} \\ \hline &\color{blue}{12}&\color{blue}{2}&\color{blue}{-19}&\color{orangered}{20} \end{array} $$The solution is:
$$ \frac{ 12x^{3}+26x^{2}-15x-18 }{ x+2 } = \color{blue}{12x^{2}+2x-19} ~+~ \frac{ \color{red}{ 20 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&26&-15&-18\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 12 }&26&-15&-18\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&26&-15&-18\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 26 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}-2&12&\color{orangered}{ 26 }&-15&-18\\& & \color{orangered}{-24} & & \\ \hline &12&\color{orangered}{2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&26&-15&-18\\& & -24& \color{blue}{-4} & \\ \hline &12&\color{blue}{2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}-2&12&26&\color{orangered}{ -15 }&-18\\& & -24& \color{orangered}{-4} & \\ \hline &12&2&\color{orangered}{-19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&26&-15&-18\\& & -24& -4& \color{blue}{38} \\ \hline &12&2&\color{blue}{-19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 38 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}-2&12&26&-15&\color{orangered}{ -18 }\\& & -24& -4& \color{orangered}{38} \\ \hline &\color{blue}{12}&\color{blue}{2}&\color{blue}{-19}&\color{orangered}{20} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}+2x-19 } $ with a remainder of $ \color{red}{ 20 } $.