Find remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ 2x+1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&1&-3&-14&12&40\\& & -1& 4& 10& \color{black}{-22} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-10}&\color{blue}{22}&\color{orangered}{18} \end{array} $$The remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x+1 $ is $ \, \color{red}{ 18 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-3&-14&12&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 1 }&-3&-14&12&40\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-3&-14&12&40\\& & \color{blue}{-1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-1&1&\color{orangered}{ -3 }&-14&12&40\\& & \color{orangered}{-1} & & & \\ \hline &1&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-3&-14&12&40\\& & -1& \color{blue}{4} & & \\ \hline &1&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 4 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-1&1&-3&\color{orangered}{ -14 }&12&40\\& & -1& \color{orangered}{4} & & \\ \hline &1&-4&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-3&-14&12&40\\& & -1& 4& \color{blue}{10} & \\ \hline &1&-4&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 10 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}-1&1&-3&-14&\color{orangered}{ 12 }&40\\& & -1& 4& \color{orangered}{10} & \\ \hline &1&-4&-10&\color{orangered}{22}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 22 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&1&-3&-14&12&40\\& & -1& 4& 10& \color{blue}{-22} \\ \hline &1&-4&-10&\color{blue}{22}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}-1&1&-3&-14&12&\color{orangered}{ 40 }\\& & -1& 4& 10& \color{orangered}{-22} \\ \hline &\color{blue}{1}&\color{blue}{-4}&\color{blue}{-10}&\color{blue}{22}&\color{orangered}{18} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 18 }\right) $.