Find remainder when $ 2x^{3}+6x^{2}+2x-8 $ is divided by $ x-3 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&6&2&-8\\& & 6& 36& \color{black}{114} \\ \hline &\color{blue}{2}&\color{blue}{12}&\color{blue}{38}&\color{orangered}{106} \end{array} $$The remainder when $ 2x^{3}+6x^{2}+2x-8 $ is divided by $ x-3 $ is $ \, \color{red}{ 106 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&6&2&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&6&2&-8\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&6&2&-8\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 6 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ 6 }&2&-8\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 12 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&6&2&-8\\& & 6& \color{blue}{36} & \\ \hline &2&\color{blue}{12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 36 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrr}3&2&6&\color{orangered}{ 2 }&-8\\& & 6& \color{orangered}{36} & \\ \hline &2&12&\color{orangered}{38}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 38 } = \color{blue}{ 114 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&6&2&-8\\& & 6& 36& \color{blue}{114} \\ \hline &2&12&\color{blue}{38}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 114 } = \color{orangered}{ 106 } $
$$ \begin{array}{c|rrrr}3&2&6&2&\color{orangered}{ -8 }\\& & 6& 36& \color{orangered}{114} \\ \hline &\color{blue}{2}&\color{blue}{12}&\color{blue}{38}&\color{orangered}{106} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 106 }\right) $.