Find remainder when $ 3x^{3}-4x^{2}-3x+5 $ is divided by $ x-4 $.
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&3&-4&-3&5\\& & 12& 32& \color{black}{116} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{29}&\color{orangered}{121} \end{array} $$The remainder when $ 3x^{3}-4x^{2}-3x+5 $ is divided by $ x-4 $ is $ \, \color{red}{ 121 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-4&-3&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 3 }&-4&-3&5\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 3 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-4&-3&5\\& & \color{blue}{12} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 12 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}4&3&\color{orangered}{ -4 }&-3&5\\& & \color{orangered}{12} & & \\ \hline &3&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 8 } = \color{blue}{ 32 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-4&-3&5\\& & 12& \color{blue}{32} & \\ \hline &3&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 32 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrr}4&3&-4&\color{orangered}{ -3 }&5\\& & 12& \color{orangered}{32} & \\ \hline &3&8&\color{orangered}{29}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 29 } = \color{blue}{ 116 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&3&-4&-3&5\\& & 12& 32& \color{blue}{116} \\ \hline &3&8&\color{blue}{29}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 116 } = \color{orangered}{ 121 } $
$$ \begin{array}{c|rrrr}4&3&-4&-3&\color{orangered}{ 5 }\\& & 12& 32& \color{orangered}{116} \\ \hline &\color{blue}{3}&\color{blue}{8}&\color{blue}{29}&\color{orangered}{121} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 121 }\right) $.