Find remainder when $ 2x^{5}-x^{3}+3x^{2}-2x+7 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&2&0&-1&3&-2&7\\& & 6& 18& 51& 162& \color{black}{480} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{17}&\color{blue}{54}&\color{blue}{160}&\color{orangered}{487} \end{array} $$The remainder when $ 2x^{5}-x^{3}+3x^{2}-2x+7 $ is divided by $ x-3 $ is $ \, \color{red}{ 487 } $.
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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 2 }&0&-1&3&-2&7\\& & & & & & \\ \hline &\color{orangered}{2}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & \color{blue}{6} & & & & \\ \hline &\color{blue}{2}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 6 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}3&2&\color{orangered}{ 0 }&-1&3&-2&7\\& & \color{orangered}{6} & & & & \\ \hline &2&\color{orangered}{6}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & 6& \color{blue}{18} & & & \\ \hline &2&\color{blue}{6}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 18 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrrr}3&2&0&\color{orangered}{ -1 }&3&-2&7\\& & 6& \color{orangered}{18} & & & \\ \hline &2&6&\color{orangered}{17}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 17 } = \color{blue}{ 51 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & 6& 18& \color{blue}{51} & & \\ \hline &2&6&\color{blue}{17}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 51 } = \color{orangered}{ 54 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&\color{orangered}{ 3 }&-2&7\\& & 6& 18& \color{orangered}{51} & & \\ \hline &2&6&17&\color{orangered}{54}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 54 } = \color{blue}{ 162 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & 6& 18& 51& \color{blue}{162} & \\ \hline &2&6&17&\color{blue}{54}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 162 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&3&\color{orangered}{ -2 }&7\\& & 6& 18& 51& \color{orangered}{162} & \\ \hline &2&6&17&54&\color{orangered}{160}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 160 } = \color{blue}{ 480 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&2&0&-1&3&-2&7\\& & 6& 18& 51& 162& \color{blue}{480} \\ \hline &2&6&17&54&\color{blue}{160}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 480 } = \color{orangered}{ 487 } $
$$ \begin{array}{c|rrrrrr}3&2&0&-1&3&-2&\color{orangered}{ 7 }\\& & 6& 18& 51& 162& \color{orangered}{480} \\ \hline &\color{blue}{2}&\color{blue}{6}&\color{blue}{17}&\color{blue}{54}&\color{blue}{160}&\color{orangered}{487} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 487 }\right) $.