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