The synthetic division table is:
$$ \begin{array}{c|rrrr}0&2&8&-6&10\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-6}&\color{orangered}{10} \end{array} $$The remainder when $ 2x^{3}+8x^{2}-6x+10 $ is divided by $ x $ is $ \, \color{red}{ 10 } $.
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.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&8&-6&10\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&8&-6&10\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&8&-6&10\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 0 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ 8 }&-6&10\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 8 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&8&-6&10\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 0 } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}0&2&8&\color{orangered}{ -6 }&10\\& & 0& \color{orangered}{0} & \\ \hline &2&8&\color{orangered}{-6}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&8&-6&10\\& & 0& 0& \color{blue}{0} \\ \hline &2&8&\color{blue}{-6}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 0 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}0&2&8&-6&\color{orangered}{ 10 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{8}&\color{blue}{-6}&\color{orangered}{10} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 10 }\right) $.