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