Divide using synthetic division $$ ( 14x^{4}-5x^{3}-11x^{2}+8 ) \div ( 2x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&14&-5&-11&0&8\\& & 14& 9& -2& \color{black}{-2} \\ \hline &\color{blue}{14}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 14x^{4}-5x^{3}-11x^{2}+8 }{ x-1 } = \color{blue}{14x^{3}+9x^{2}-2x-2} ~+~ \frac{ \color{red}{ 6 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&14&-5&-11&0&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 14 }&-5&-11&0&8\\& & & & & \\ \hline &\color{orangered}{14}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 14 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&14&-5&-11&0&8\\& & \color{blue}{14} & & & \\ \hline &\color{blue}{14}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 14 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&14&\color{orangered}{ -5 }&-11&0&8\\& & \color{orangered}{14} & & & \\ \hline &14&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&14&-5&-11&0&8\\& & 14& \color{blue}{9} & & \\ \hline &14&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 9 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&14&-5&\color{orangered}{ -11 }&0&8\\& & 14& \color{orangered}{9} & & \\ \hline &14&9&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&14&-5&-11&0&8\\& & 14& 9& \color{blue}{-2} & \\ \hline &14&9&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&14&-5&-11&\color{orangered}{ 0 }&8\\& & 14& 9& \color{orangered}{-2} & \\ \hline &14&9&-2&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&14&-5&-11&0&8\\& & 14& 9& -2& \color{blue}{-2} \\ \hline &14&9&-2&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}1&14&-5&-11&0&\color{orangered}{ 8 }\\& & 14& 9& -2& \color{orangered}{-2} \\ \hline &\color{blue}{14}&\color{blue}{9}&\color{blue}{-2}&\color{blue}{-2}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 14x^{3}+9x^{2}-2x-2 } $ with a remainder of $ \color{red}{ 6 } $.