Find remainder when $ 3x^{3}+7x^{2}+5x+2 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&7&5&2\\& & 6& 26& \color{black}{62} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{31}&\color{orangered}{64} \end{array} $$The remainder when $ 3x^{3}+7x^{2}+5x+2 $ is divided by $ x-2 $ is $ \, \color{red}{ 64 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&7&5&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&7&5&2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&7&5&2\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 6 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ 7 }&5&2\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 13 } = \color{blue}{ 26 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&7&5&2\\& & 6& \color{blue}{26} & \\ \hline &3&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 26 } = \color{orangered}{ 31 } $
$$ \begin{array}{c|rrrr}2&3&7&\color{orangered}{ 5 }&2\\& & 6& \color{orangered}{26} & \\ \hline &3&13&\color{orangered}{31}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 31 } = \color{blue}{ 62 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&7&5&2\\& & 6& 26& \color{blue}{62} \\ \hline &3&13&\color{blue}{31}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 62 } = \color{orangered}{ 64 } $
$$ \begin{array}{c|rrrr}2&3&7&5&\color{orangered}{ 2 }\\& & 6& 26& \color{orangered}{62} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{31}&\color{orangered}{64} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 64 }\right) $.