Divide using synthetic division $$ ( x^{3}+6x^{2}-11x-5 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&6&-11&-5\\& & 2& 16& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{5}&\color{orangered}{5} \end{array} $$The solution is:
$$ \frac{ x^{3}+6x^{2}-11x-5 }{ x-2 } = \color{blue}{x^{2}+8x+5} ~+~ \frac{ \color{red}{ 5 } }{ x-2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&-11&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&6&-11&-5\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&-11&-5\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 2 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 6 }&-11&-5\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&-11&-5\\& & 2& \color{blue}{16} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 16 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&1&6&\color{orangered}{ -11 }&-5\\& & 2& \color{orangered}{16} & \\ \hline &1&8&\color{orangered}{5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&6&-11&-5\\& & 2& 16& \color{blue}{10} \\ \hline &1&8&\color{blue}{5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 10 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}2&1&6&-11&\color{orangered}{ -5 }\\& & 2& 16& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{5}&\color{orangered}{5} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+8x+5 } $ with a remainder of $ \color{red}{ 5 } $.