Find remainder when $ x^{4}+4x^{3}+6x^{2}-3x-7 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&4&6&-3&-7\\& & 2& 12& 36& \color{black}{66} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{18}&\color{blue}{33}&\color{orangered}{59} \end{array} $$The remainder when $ x^{4}+4x^{3}+6x^{2}-3x-7 $ is divided by $ x-2 $ is $ \, \color{red}{ 59 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&6&-3&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&4&6&-3&-7\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&6&-3&-7\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 4 }&6&-3&-7\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&6&-3&-7\\& & 2& \color{blue}{12} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 12 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrrr}2&1&4&\color{orangered}{ 6 }&-3&-7\\& & 2& \color{orangered}{12} & & \\ \hline &1&6&\color{orangered}{18}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&6&-3&-7\\& & 2& 12& \color{blue}{36} & \\ \hline &1&6&\color{blue}{18}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 36 } = \color{orangered}{ 33 } $
$$ \begin{array}{c|rrrrr}2&1&4&6&\color{orangered}{ -3 }&-7\\& & 2& 12& \color{orangered}{36} & \\ \hline &1&6&18&\color{orangered}{33}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 33 } = \color{blue}{ 66 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&6&-3&-7\\& & 2& 12& 36& \color{blue}{66} \\ \hline &1&6&18&\color{blue}{33}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 66 } = \color{orangered}{ 59 } $
$$ \begin{array}{c|rrrrr}2&1&4&6&-3&\color{orangered}{ -7 }\\& & 2& 12& 36& \color{orangered}{66} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{18}&\color{blue}{33}&\color{orangered}{59} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 59 }\right) $.