Find remainder when $ x^{4}+x^{3}+x^{2}+3x+5 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&1&1&3&5\\& & 4& 20& 84& \color{black}{348} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{21}&\color{blue}{87}&\color{orangered}{353} \end{array} $$The remainder when $ x^{4}+x^{3}+x^{2}+3x+5 $ is divided by $ x-4 $ is $ \, \color{red}{ 353 } $.
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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&1&3&5\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&1&1&3&5\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&1&3&5\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ 1 }&1&3&5\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&1&3&5\\& & 4& \color{blue}{20} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 20 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrrr}4&1&1&\color{orangered}{ 1 }&3&5\\& & 4& \color{orangered}{20} & & \\ \hline &1&5&\color{orangered}{21}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 21 } = \color{blue}{ 84 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&1&3&5\\& & 4& 20& \color{blue}{84} & \\ \hline &1&5&\color{blue}{21}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 84 } = \color{orangered}{ 87 } $
$$ \begin{array}{c|rrrrr}4&1&1&1&\color{orangered}{ 3 }&5\\& & 4& 20& \color{orangered}{84} & \\ \hline &1&5&21&\color{orangered}{87}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 87 } = \color{blue}{ 348 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&1&1&3&5\\& & 4& 20& 84& \color{blue}{348} \\ \hline &1&5&21&\color{blue}{87}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 348 } = \color{orangered}{ 353 } $
$$ \begin{array}{c|rrrrr}4&1&1&1&3&\color{orangered}{ 5 }\\& & 4& 20& 84& \color{orangered}{348} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{21}&\color{blue}{87}&\color{orangered}{353} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 353 }\right) $.