Check if $ x-2 $ is a factor of $ 5x^{4}-x^{2}+1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&0&-1&0&1\\& & 10& 20& 38& \color{black}{76} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{19}&\color{blue}{38}&\color{orangered}{77} \end{array} $$Because the remainder $ \left( \color{red}{ 77 } \right) $ is not zero, we conclude that the $ x-2 $ is not a factor of $ 5x^{4}-x^{2}+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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-1&0&1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&0&-1&0&1\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-1&0&1\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 10 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ 0 }&-1&0&1\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 10 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-1&0&1\\& & 10& \color{blue}{20} & & \\ \hline &5&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 20 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrrr}2&5&0&\color{orangered}{ -1 }&0&1\\& & 10& \color{orangered}{20} & & \\ \hline &5&10&\color{orangered}{19}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 19 } = \color{blue}{ 38 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-1&0&1\\& & 10& 20& \color{blue}{38} & \\ \hline &5&10&\color{blue}{19}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 38 } = \color{orangered}{ 38 } $
$$ \begin{array}{c|rrrrr}2&5&0&-1&\color{orangered}{ 0 }&1\\& & 10& 20& \color{orangered}{38} & \\ \hline &5&10&19&\color{orangered}{38}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 38 } = \color{blue}{ 76 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&0&-1&0&1\\& & 10& 20& 38& \color{blue}{76} \\ \hline &5&10&19&\color{blue}{38}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 76 } = \color{orangered}{ 77 } $
$$ \begin{array}{c|rrrrr}2&5&0&-1&0&\color{orangered}{ 1 }\\& & 10& 20& 38& \color{orangered}{76} \\ \hline &\color{blue}{5}&\color{blue}{10}&\color{blue}{19}&\color{blue}{38}&\color{orangered}{77} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 77 }\right)$.