Check if $ x-1 $ is a factor of $ -3x^{4}-2x^{3}+3x^{2}+x-2 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&-3&-2&3&1&-2\\& & -3& -5& -2& \color{black}{-1} \\ \hline &\color{blue}{-3}&\color{blue}{-5}&\color{blue}{-2}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$Because the remainder $ \left( \color{red}{ -3 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ -3x^{4}-2x^{3}+3x^{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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-3&-2&3&1&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ -3 }&-2&3&1&-2\\& & & & & \\ \hline &\color{orangered}{-3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-3&-2&3&1&-2\\& & \color{blue}{-3} & & & \\ \hline &\color{blue}{-3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&-3&\color{orangered}{ -2 }&3&1&-2\\& & \color{orangered}{-3} & & & \\ \hline &-3&\color{orangered}{-5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-3&-2&3&1&-2\\& & -3& \color{blue}{-5} & & \\ \hline &-3&\color{blue}{-5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}1&-3&-2&\color{orangered}{ 3 }&1&-2\\& & -3& \color{orangered}{-5} & & \\ \hline &-3&-5&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-3&-2&3&1&-2\\& & -3& -5& \color{blue}{-2} & \\ \hline &-3&-5&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}1&-3&-2&3&\color{orangered}{ 1 }&-2\\& & -3& -5& \color{orangered}{-2} & \\ \hline &-3&-5&-2&\color{orangered}{-1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&-3&-2&3&1&-2\\& & -3& -5& -2& \color{blue}{-1} \\ \hline &-3&-5&-2&\color{blue}{-1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&-3&-2&3&1&\color{orangered}{ -2 }\\& & -3& -5& -2& \color{orangered}{-1} \\ \hline &\color{blue}{-3}&\color{blue}{-5}&\color{blue}{-2}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -3 }\right)$.