Divide using synthetic division $$ ( 3x^{3}+11x^{2}-2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&11&0&-2\\& & -6& -10& \color{black}{20} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-10}&\color{orangered}{18} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+11x^{2}-2 }{ x+2 } = \color{blue}{3x^{2}+5x-10} ~+~ \frac{ \color{red}{ 18 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&0&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&11&0&-2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&0&-2\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 11 }&0&-2\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&0&-2\\& & -6& \color{blue}{-10} & \\ \hline &3&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-2&3&11&\color{orangered}{ 0 }&-2\\& & -6& \color{orangered}{-10} & \\ \hline &3&5&\color{orangered}{-10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&11&0&-2\\& & -6& -10& \color{blue}{20} \\ \hline &3&5&\color{blue}{-10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 20 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}-2&3&11&0&\color{orangered}{ -2 }\\& & -6& -10& \color{orangered}{20} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{-10}&\color{orangered}{18} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+5x-10 } $ with a remainder of $ \color{red}{ 18 } $.