The synthetic division table is:
$$ \begin{array}{c|rrr}0&4&-2&2\\& & 0& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{orangered}{2} \end{array} $$The remainder when $ 4x^{2}-2x+2 $ is divided by $ x $ is $ \, \color{red}{ 2 } $.
We can find remainder using synthetic division method.
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrr}\color{blue}{0}&4&-2&2\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}0&\color{orangered}{ 4 }&-2&2\\& & & \\ \hline &\color{orangered}{4}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&4&-2&2\\& & \color{blue}{0} & \\ \hline &\color{blue}{4}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 0 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrr}0&4&\color{orangered}{ -2 }&2\\& & \color{orangered}{0} & \\ \hline &4&\color{orangered}{-2}& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{0}&4&-2&2\\& & 0& \color{blue}{0} \\ \hline &4&\color{blue}{-2}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrr}0&4&-2&\color{orangered}{ 2 }\\& & 0& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{-2}&\color{orangered}{2} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 2 }\right) $.