Check if $ x+2 $ is a factor of $ 3x^{3}+x^{2}+x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&1&1&-2\\& & -6& 10& \color{black}{-22} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{11}&\color{orangered}{-24} \end{array} $$Because the remainder $ \left( \color{red}{ -24 } \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}{ \left( -6 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 1 }&1&-2\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&1&-2\\& & -6& \color{blue}{10} & \\ \hline &3&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 10 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrr}-2&3&1&\color{orangered}{ 1 }&-2\\& & -6& \color{orangered}{10} & \\ \hline &3&-5&\color{orangered}{11}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 11 } = \color{blue}{ -22 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&1&1&-2\\& & -6& 10& \color{blue}{-22} \\ \hline &3&-5&\color{blue}{11}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -22 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrr}-2&3&1&1&\color{orangered}{ -2 }\\& & -6& 10& \color{orangered}{-22} \\ \hline &\color{blue}{3}&\color{blue}{-5}&\color{blue}{11}&\color{orangered}{-24} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -24 }\right)$.