The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&-2&3&-4\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-4} \end{array} $$Because the remainder $ \left( \color{red}{ -4 } \right) $ is not zero, we conclude that the $ x $ is not a factor of $ x^{3}-2x^{2}+3x-4$.
First we need to create a synthetic division table.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-2&3&-4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&-2&3&-4\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-2&3&-4\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ -2 }&3&-4\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{-2}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-2&3&-4\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{-2}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 0 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}0&1&-2&\color{orangered}{ 3 }&-4\\& & 0& \color{orangered}{0} & \\ \hline &1&-2&\color{orangered}{3}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 3 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&-2&3&-4\\& & 0& 0& \color{blue}{0} \\ \hline &1&-2&\color{blue}{3}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 0 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}0&1&-2&3&\color{orangered}{ -4 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{-2}&\color{blue}{3}&\color{orangered}{-4} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -4 }\right)$.