Find remainder when $ 3x^{4}+8 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&3&0&0&0&8\\& & 9& 27& 81& \color{black}{243} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{27}&\color{blue}{81}&\color{orangered}{251} \end{array} $$The remainder when $ 3x^{4}+8 $ is divided by $ x-3 $ is $ \, \color{red}{ 251 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&0&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 3 }&0&0&0&8\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&0&8\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}3&3&\color{orangered}{ 0 }&0&0&8\\& & \color{orangered}{9} & & & \\ \hline &3&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&0&8\\& & 9& \color{blue}{27} & & \\ \hline &3&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 27 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}3&3&0&\color{orangered}{ 0 }&0&8\\& & 9& \color{orangered}{27} & & \\ \hline &3&9&\color{orangered}{27}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 27 } = \color{blue}{ 81 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&0&8\\& & 9& 27& \color{blue}{81} & \\ \hline &3&9&\color{blue}{27}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 81 } = \color{orangered}{ 81 } $
$$ \begin{array}{c|rrrrr}3&3&0&0&\color{orangered}{ 0 }&8\\& & 9& 27& \color{orangered}{81} & \\ \hline &3&9&27&\color{orangered}{81}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 81 } = \color{blue}{ 243 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&3&0&0&0&8\\& & 9& 27& 81& \color{blue}{243} \\ \hline &3&9&27&\color{blue}{81}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 243 } = \color{orangered}{ 251 } $
$$ \begin{array}{c|rrrrr}3&3&0&0&0&\color{orangered}{ 8 }\\& & 9& 27& 81& \color{orangered}{243} \\ \hline &\color{blue}{3}&\color{blue}{9}&\color{blue}{27}&\color{blue}{81}&\color{orangered}{251} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 251 }\right) $.