Find remainder when $ 3x^{5}-5x^{4}-2x^{3}+2x^{2}+4 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}2&3&-5&-2&2&0&4\\& & 6& 2& 0& 4& \color{black}{8} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{0}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{12} \end{array} $$The remainder when $ 3x^{5}-5x^{4}-2x^{3}+2x^{2}+4 $ is divided by $ x-2 $ is $ \, \color{red}{ 12 } $.
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.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}2&\color{orangered}{ 3 }&-5&-2&2&0&4\\& & & & & & \\ \hline &\color{orangered}{3}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{3}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 6 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}2&3&\color{orangered}{ -5 }&-2&2&0&4\\& & \color{orangered}{6} & & & & \\ \hline &3&\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & 6& \color{blue}{2} & & & \\ \hline &3&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 2 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}2&3&-5&\color{orangered}{ -2 }&2&0&4\\& & 6& \color{orangered}{2} & & & \\ \hline &3&1&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & 6& 2& \color{blue}{0} & & \\ \hline &3&1&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrrr}2&3&-5&-2&\color{orangered}{ 2 }&0&4\\& & 6& 2& \color{orangered}{0} & & \\ \hline &3&1&0&\color{orangered}{2}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & 6& 2& 0& \color{blue}{4} & \\ \hline &3&1&0&\color{blue}{2}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}2&3&-5&-2&2&\color{orangered}{ 0 }&4\\& & 6& 2& 0& \color{orangered}{4} & \\ \hline &3&1&0&2&\color{orangered}{4}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{2}&3&-5&-2&2&0&4\\& & 6& 2& 0& 4& \color{blue}{8} \\ \hline &3&1&0&2&\color{blue}{4}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 8 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrrr}2&3&-5&-2&2&0&\color{orangered}{ 4 }\\& & 6& 2& 0& 4& \color{orangered}{8} \\ \hline &\color{blue}{3}&\color{blue}{1}&\color{blue}{0}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 12 }\right) $.