Find remainder when $ 4x^{4}+3x-4 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&4&0&0&3&-4\\& & 12& 36& 108& \color{black}{333} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{36}&\color{blue}{111}&\color{orangered}{329} \end{array} $$The remainder when $ 4x^{4}+3x-4 $ is divided by $ x-3 $ is $ \, \color{red}{ 329 } $.
Explanation
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&0&3&-4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 4 }&0&0&3&-4\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&0&3&-4\\& & \color{blue}{12} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 12 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}3&4&\color{orangered}{ 0 }&0&3&-4\\& & \color{orangered}{12} & & & \\ \hline &4&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&0&3&-4\\& & 12& \color{blue}{36} & & \\ \hline &4&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 36 } = \color{orangered}{ 36 } $
$$ \begin{array}{c|rrrrr}3&4&0&\color{orangered}{ 0 }&3&-4\\& & 12& \color{orangered}{36} & & \\ \hline &4&12&\color{orangered}{36}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 36 } = \color{blue}{ 108 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&0&3&-4\\& & 12& 36& \color{blue}{108} & \\ \hline &4&12&\color{blue}{36}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 108 } = \color{orangered}{ 111 } $
$$ \begin{array}{c|rrrrr}3&4&0&0&\color{orangered}{ 3 }&-4\\& & 12& 36& \color{orangered}{108} & \\ \hline &4&12&36&\color{orangered}{111}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 111 } = \color{blue}{ 333 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&4&0&0&3&-4\\& & 12& 36& 108& \color{blue}{333} \\ \hline &4&12&36&\color{blue}{111}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 333 } = \color{orangered}{ 329 } $
$$ \begin{array}{c|rrrrr}3&4&0&0&3&\color{orangered}{ -4 }\\& & 12& 36& 108& \color{orangered}{333} \\ \hline &\color{blue}{4}&\color{blue}{12}&\color{blue}{36}&\color{blue}{111}&\color{orangered}{329} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 329 }\right) $.