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

The synthetic division table is:

$$ \begin{array}{c|rrrr}1&3&1&1&-2\\& & 3& 4& \color{black}{5} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{5}&\color{orangered}{3} \end{array} $$

Because the remainder $ \left( \color{red}{ 3 } \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}{ 3 } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrr}1&3&\color{orangered}{ 1 }&1&-2\\& & \color{orangered}{3} & & \\ \hline &3&\color{orangered}{4}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&1&-2\\& & 3& \color{blue}{4} & \\ \hline &3&\color{blue}{4}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 4 } = \color{orangered}{ 5 } $

$$ \begin{array}{c|rrrr}1&3&1&\color{orangered}{ 1 }&-2\\& & 3& \color{orangered}{4} & \\ \hline &3&4&\color{orangered}{5}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 5 } = \color{blue}{ 5 } $.

$$ \begin{array}{c|rrrr}\color{blue}{1}&3&1&1&-2\\& & 3& 4& \color{blue}{5} \\ \hline &3&4&\color{blue}{5}& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&3&1&1&\color{orangered}{ -2 }\\& & 3& 4& \color{orangered}{5} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{5}&\color{orangered}{3} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 3 }\right)$.

Synthetic Division Calculator