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