Find remainder when $ x^{3}-x+4 $ is divided by $ x-8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&1&0&-1&4\\& & 8& 64& \color{black}{504} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{63}&\color{orangered}{508} \end{array} $$The remainder when $ x^{3}-x+4 $ is divided by $ x-8 $ is $ \, \color{red}{ 508 } $.
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 -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-1&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 1 }&0&-1&4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-1&4\\& & \color{blue}{8} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 8 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}8&1&\color{orangered}{ 0 }&-1&4\\& & \color{orangered}{8} & & \\ \hline &1&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 8 } = \color{blue}{ 64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-1&4\\& & 8& \color{blue}{64} & \\ \hline &1&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 64 } = \color{orangered}{ 63 } $
$$ \begin{array}{c|rrrr}8&1&0&\color{orangered}{ -1 }&4\\& & 8& \color{orangered}{64} & \\ \hline &1&8&\color{orangered}{63}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 63 } = \color{blue}{ 504 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&1&0&-1&4\\& & 8& 64& \color{blue}{504} \\ \hline &1&8&\color{blue}{63}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 504 } = \color{orangered}{ 508 } $
$$ \begin{array}{c|rrrr}8&1&0&-1&\color{orangered}{ 4 }\\& & 8& 64& \color{orangered}{504} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{63}&\color{orangered}{508} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 508 }\right) $.