Find remainder when $ x^{4}+4x^{3}-x^{2}-16x-14 $ is divided by $ 2x-1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&4&-1&-16&-14\\& & 1& 5& 4& \color{black}{-12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{4}&\color{blue}{-12}&\color{orangered}{-26} \end{array} $$The remainder when $ x^{4}+4x^{3}-x^{2}-16x-14 $ is divided by $ x-1 $ is $ \, \color{red}{ -26 } $.
We can find remainder using synthetic division method.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-1&-16&-14\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&4&-1&-16&-14\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-1&-16&-14\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 1 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 4 }&-1&-16&-14\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-1&-16&-14\\& & 1& \color{blue}{5} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 5 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}1&1&4&\color{orangered}{ -1 }&-16&-14\\& & 1& \color{orangered}{5} & & \\ \hline &1&5&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-1&-16&-14\\& & 1& 5& \color{blue}{4} & \\ \hline &1&5&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 4 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}1&1&4&-1&\color{orangered}{ -16 }&-14\\& & 1& 5& \color{orangered}{4} & \\ \hline &1&5&4&\color{orangered}{-12}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&4&-1&-16&-14\\& & 1& 5& 4& \color{blue}{-12} \\ \hline &1&5&4&\color{blue}{-12}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -26 } $
$$ \begin{array}{c|rrrrr}1&1&4&-1&-16&\color{orangered}{ -14 }\\& & 1& 5& 4& \color{orangered}{-12} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{4}&\color{blue}{-12}&\color{orangered}{-26} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -26 }\right) $.