Divide using synthetic division $$ ( 6x^{4}+11x^{3}-7x^{2}-8x+11 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&6&11&-7&-8&11\\& & 6& 17& 10& \color{black}{2} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{10}&\color{blue}{2}&\color{orangered}{13} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+11x^{3}-7x^{2}-8x+11 }{ x-1 } = \color{blue}{6x^{3}+17x^{2}+10x+2} ~+~ \frac{ \color{red}{ 13 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&11&-7&-8&11\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 6 }&11&-7&-8&11\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 6 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&11&-7&-8&11\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 6 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}1&6&\color{orangered}{ 11 }&-7&-8&11\\& & \color{orangered}{6} & & & \\ \hline &6&\color{orangered}{17}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&11&-7&-8&11\\& & 6& \color{blue}{17} & & \\ \hline &6&\color{blue}{17}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 17 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}1&6&11&\color{orangered}{ -7 }&-8&11\\& & 6& \color{orangered}{17} & & \\ \hline &6&17&\color{orangered}{10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 10 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&11&-7&-8&11\\& & 6& 17& \color{blue}{10} & \\ \hline &6&17&\color{blue}{10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 10 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}1&6&11&-7&\color{orangered}{ -8 }&11\\& & 6& 17& \color{orangered}{10} & \\ \hline &6&17&10&\color{orangered}{2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&6&11&-7&-8&11\\& & 6& 17& 10& \color{blue}{2} \\ \hline &6&17&10&\color{blue}{2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 11 } + \color{orangered}{ 2 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}1&6&11&-7&-8&\color{orangered}{ 11 }\\& & 6& 17& 10& \color{orangered}{2} \\ \hline &\color{blue}{6}&\color{blue}{17}&\color{blue}{10}&\color{blue}{2}&\color{orangered}{13} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}+17x^{2}+10x+2 } $ with a remainder of $ \color{red}{ 13 } $.