Find remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x-2 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&-3&-14&12&40\\& & 2& -2& -32& \color{black}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-16}&\color{blue}{-20}&\color{orangered}{0} \end{array} $$The remainder when $ x^{4}-3x^{3}-14x^{2}+12x+40 $ is divided by $ x-2 $ is $ \, \color{red}{ 0 } $.
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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&-14&12&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&-3&-14&12&40\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&-14&12&40\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 2 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ -3 }&-14&12&40\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&-14&12&40\\& & 2& \color{blue}{-2} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -16 } $
$$ \begin{array}{c|rrrrr}2&1&-3&\color{orangered}{ -14 }&12&40\\& & 2& \color{orangered}{-2} & & \\ \hline &1&-1&\color{orangered}{-16}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -16 \right) } = \color{blue}{ -32 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&-14&12&40\\& & 2& -2& \color{blue}{-32} & \\ \hline &1&-1&\color{blue}{-16}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -32 \right) } = \color{orangered}{ -20 } $
$$ \begin{array}{c|rrrrr}2&1&-3&-14&\color{orangered}{ 12 }&40\\& & 2& -2& \color{orangered}{-32} & \\ \hline &1&-1&-16&\color{orangered}{-20}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -20 \right) } = \color{blue}{ -40 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&-3&-14&12&40\\& & 2& -2& -32& \color{blue}{-40} \\ \hline &1&-1&-16&\color{blue}{-20}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ \left( -40 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&1&-3&-14&12&\color{orangered}{ 40 }\\& & 2& -2& -32& \color{orangered}{-40} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{-16}&\color{blue}{-20}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.