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