Check if $ x+1 $ is a factor of $ 3x^{3}+x^{2}+x-2 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}-1&3&1&1&-2\\& & -3& 2& \color{black}{-3} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-5} \end{array} $$

Because the remainder $ \left( \color{red}{ -5 } \right) $ is not zero, we conclude that the $ x+1 $ is not a factor of $ 3x^{3}+x^{2}+x-2$.

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 + 1 = 0 $ ( $ x = \color{blue}{ -1 } $ ) at the left.

$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&1&1&-2\\& & & & \\ \hline &&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 3 }&1&1&-2\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.

$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&1&1&-2\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{3}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $

$$ \begin{array}{c|rrrr}-1&3&\color{orangered}{ 1 }&1&-2\\& & \color{orangered}{-3} & & \\ \hline &3&\color{orangered}{-2}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.

$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&1&1&-2\\& & -3& \color{blue}{2} & \\ \hline &3&\color{blue}{-2}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrr}-1&3&1&\color{orangered}{ 1 }&-2\\& & -3& \color{orangered}{2} & \\ \hline &3&-2&\color{orangered}{3}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.

$$ \begin{array}{c|rrrr}\color{blue}{-1}&3&1&1&-2\\& & -3& 2& \color{blue}{-3} \\ \hline &3&-2&\color{blue}{3}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -5 } $

$$ \begin{array}{c|rrrr}-1&3&1&1&\color{orangered}{ -2 }\\& & -3& 2& \color{orangered}{-3} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-5} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ -5 }\right)$.

Synthetic Division Calculator