Find remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&16&60&40\\& & 1& 17& \color{black}{77} \\ \hline &\color{blue}{1}&\color{blue}{17}&\color{blue}{77}&\color{orangered}{117} \end{array} $$The remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x-1 $ is $ \, \color{red}{ 117 } $.
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 -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&16&60&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&16&60&40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&16&60&40\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 1 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 16 }&60&40\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{17}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&16&60&40\\& & 1& \color{blue}{17} & \\ \hline &1&\color{blue}{17}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ 17 } = \color{orangered}{ 77 } $
$$ \begin{array}{c|rrrr}1&1&16&\color{orangered}{ 60 }&40\\& & 1& \color{orangered}{17} & \\ \hline &1&17&\color{orangered}{77}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 77 } = \color{blue}{ 77 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&16&60&40\\& & 1& 17& \color{blue}{77} \\ \hline &1&17&\color{blue}{77}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 77 } = \color{orangered}{ 117 } $
$$ \begin{array}{c|rrrr}1&1&16&60&\color{orangered}{ 40 }\\& & 1& 17& \color{orangered}{77} \\ \hline &\color{blue}{1}&\color{blue}{17}&\color{blue}{77}&\color{orangered}{117} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 117 }\right) $.