Find remainder when $ 5x^{3}+x-3 $ is divided by $ x-4 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&5&0&1&-3\\& & 20& 80& \color{black}{324} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{81}&\color{orangered}{321} \end{array} $$The remainder when $ 5x^{3}+x-3 $ is divided by $ x-4 $ is $ \, \color{red}{ 321 } $.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&0&1&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 5 }&0&1&-3\\& & & & \\ \hline &\color{orangered}{5}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&0&1&-3\\& & \color{blue}{20} & & \\ \hline &\color{blue}{5}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 20 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrr}4&5&\color{orangered}{ 0 }&1&-3\\& & \color{orangered}{20} & & \\ \hline &5&\color{orangered}{20}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 20 } = \color{blue}{ 80 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&0&1&-3\\& & 20& \color{blue}{80} & \\ \hline &5&\color{blue}{20}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 80 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrr}4&5&0&\color{orangered}{ 1 }&-3\\& & 20& \color{orangered}{80} & \\ \hline &5&20&\color{orangered}{81}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 81 } = \color{blue}{ 324 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&5&0&1&-3\\& & 20& 80& \color{blue}{324} \\ \hline &5&20&\color{blue}{81}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 324 } = \color{orangered}{ 321 } $
$$ \begin{array}{c|rrrr}4&5&0&1&\color{orangered}{ -3 }\\& & 20& 80& \color{orangered}{324} \\ \hline &\color{blue}{5}&\color{blue}{20}&\color{blue}{81}&\color{orangered}{321} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 321 }\right) $.