Divide using synthetic division $$ ( x^{3}+11x^{2}-10x+6 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&1&11&-10&6\\& & -2& -18& \color{black}{56} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-28}&\color{orangered}{62} \end{array} $$The solution is:
$$ \frac{ x^{3}+11x^{2}-10x+6 }{ x+2 } = \color{blue}{x^{2}+9x-28} ~+~ \frac{ \color{red}{ 62 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&11&-10&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 1 }&11&-10&6\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&11&-10&6\\& & \color{blue}{-2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}-2&1&\color{orangered}{ 11 }&-10&6\\& & \color{orangered}{-2} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 9 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&11&-10&6\\& & -2& \color{blue}{-18} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -28 } $
$$ \begin{array}{c|rrrr}-2&1&11&\color{orangered}{ -10 }&6\\& & -2& \color{orangered}{-18} & \\ \hline &1&9&\color{orangered}{-28}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -28 \right) } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&1&11&-10&6\\& & -2& -18& \color{blue}{56} \\ \hline &1&9&\color{blue}{-28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 56 } = \color{orangered}{ 62 } $
$$ \begin{array}{c|rrrr}-2&1&11&-10&\color{orangered}{ 6 }\\& & -2& -18& \color{orangered}{56} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{-28}&\color{orangered}{62} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x-28 } $ with a remainder of $ \color{red}{ 62 } $.