Check if $ x-2 $ is a factor of $ 3x^{3}+x^{2}+x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&3&1&1&-2\\& & 6& 14& \color{black}{30} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{15}&\color{orangered}{28} \end{array} $$Because the remainder $ \left( \color{red}{ 28 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ 3x^{3}+x^{2}+x-2$.
Explanation
First we need to create a synthetic division table.
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.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&1&1&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 3 }&1&1&-2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&1&1&-2\\& & \color{blue}{6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 6 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}2&3&\color{orangered}{ 1 }&1&-2\\& & \color{orangered}{6} & & \\ \hline &3&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 7 } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&1&1&-2\\& & 6& \color{blue}{14} & \\ \hline &3&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 14 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}2&3&1&\color{orangered}{ 1 }&-2\\& & 6& \color{orangered}{14} & \\ \hline &3&7&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&3&1&1&-2\\& & 6& 14& \color{blue}{30} \\ \hline &3&7&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 30 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}2&3&1&1&\color{orangered}{ -2 }\\& & 6& 14& \color{orangered}{30} \\ \hline &\color{blue}{3}&\color{blue}{7}&\color{blue}{15}&\color{orangered}{28} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 28 }\right)$.