Find remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&1&16&60&40\\& & 2& 36& \color{black}{192} \\ \hline &\color{blue}{1}&\color{blue}{18}&\color{blue}{96}&\color{orangered}{232} \end{array} $$The remainder when $ x^{3}+16x^{2}+60x+40 $ is divided by $ x-2 $ is $ \, \color{red}{ 232 } $.
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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&16&60&40\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 1 }&16&60&40\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&16&60&40\\& & \color{blue}{2} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 2 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}2&1&\color{orangered}{ 16 }&60&40\\& & \color{orangered}{2} & & \\ \hline &1&\color{orangered}{18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&16&60&40\\& & 2& \color{blue}{36} & \\ \hline &1&\color{blue}{18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ 36 } = \color{orangered}{ 96 } $
$$ \begin{array}{c|rrrr}2&1&16&\color{orangered}{ 60 }&40\\& & 2& \color{orangered}{36} & \\ \hline &1&18&\color{orangered}{96}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 96 } = \color{blue}{ 192 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&1&16&60&40\\& & 2& 36& \color{blue}{192} \\ \hline &1&18&\color{blue}{96}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 192 } = \color{orangered}{ 232 } $
$$ \begin{array}{c|rrrr}2&1&16&60&\color{orangered}{ 40 }\\& & 2& 36& \color{orangered}{192} \\ \hline &\color{blue}{1}&\color{blue}{18}&\color{blue}{96}&\color{orangered}{232} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 232 }\right) $.