Find remainder when $ 3x^{4}-5x^{3}+5x+2 $ is divided by $ x-2 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&-5&0&5&2\\& & 6& 2& 4& \color{black}{18} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{20} \end{array} $$The remainder when $ 3x^{4}-5x^{3}+5x+2 $ is divided by $ x-2 $ is $ \, \color{red}{ 20 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-5&0&5&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&-5&0&5&2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-5&0&5&2\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ -5 }&0&5&2\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-5&0&5&2\\& & 6& \color{blue}{2} & & \\ \hline &3&\color{blue}{1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}2&3&-5&\color{orangered}{ 0 }&5&2\\& & 6& \color{orangered}{2} & & \\ \hline &3&1&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-5&0&5&2\\& & 6& 2& \color{blue}{4} & \\ \hline &3&1&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 4 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}2&3&-5&0&\color{orangered}{ 5 }&2\\& & 6& 2& \color{orangered}{4} & \\ \hline &3&1&2&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&-5&0&5&2\\& & 6& 2& 4& \color{blue}{18} \\ \hline &3&1&2&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 18 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}2&3&-5&0&5&\color{orangered}{ 2 }\\& & 6& 2& 4& \color{orangered}{18} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{2}&\color{blue}{9}&\color{orangered}{20} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 20 }\right) $.