Check if $ x-2 $ is a factor of $ 2x^{4}-7x^{2}+x-6 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&2&0&-7&1&-6\\& & 4& 8& 2& \color{black}{6} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Because the remainder equals zero, we conclude that the $ x-2 $ is a factor of the $ 2x^{4}-7x^{2}+x-6 $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&-7&1&-6\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 2 }&0&-7&1&-6\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 2 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&-7&1&-6\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 4 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}2&2&\color{orangered}{ 0 }&-7&1&-6\\& & \color{orangered}{4} & & & \\ \hline &2&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 4 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&-7&1&-6\\& & 4& \color{blue}{8} & & \\ \hline &2&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 8 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}2&2&0&\color{orangered}{ -7 }&1&-6\\& & 4& \color{orangered}{8} & & \\ \hline &2&4&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&-7&1&-6\\& & 4& 8& \color{blue}{2} & \\ \hline &2&4&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 2 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&2&0&-7&\color{orangered}{ 1 }&-6\\& & 4& 8& \color{orangered}{2} & \\ \hline &2&4&1&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&2&0&-7&1&-6\\& & 4& 8& 2& \color{blue}{6} \\ \hline &2&4&1&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 6 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}2&2&0&-7&1&\color{orangered}{ -6 }\\& & 4& 8& 2& \color{orangered}{6} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{1}&\color{blue}{3}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.