Check if $ x-1 $ is a factor of $ 9x^{4}-6x^{3}-2x^{2}+16x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&9&-6&-2&16&8\\& & 9& 3& 1& \color{black}{17} \\ \hline &\color{blue}{9}&\color{blue}{3}&\color{blue}{1}&\color{blue}{17}&\color{orangered}{25} \end{array} $$Because the remainder $ \left( \color{red}{ 25 } \right) $ is not zero, we conclude that the $ x-1 $ is not a factor of $ 9x^{4}-6x^{3}-2x^{2}+16x+8$.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-6&-2&16&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 9 }&-6&-2&16&8\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-6&-2&16&8\\& & \color{blue}{9} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 9 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}1&9&\color{orangered}{ -6 }&-2&16&8\\& & \color{orangered}{9} & & & \\ \hline &9&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-6&-2&16&8\\& & 9& \color{blue}{3} & & \\ \hline &9&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 3 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}1&9&-6&\color{orangered}{ -2 }&16&8\\& & 9& \color{orangered}{3} & & \\ \hline &9&3&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-6&-2&16&8\\& & 9& 3& \color{blue}{1} & \\ \hline &9&3&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 1 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrrr}1&9&-6&-2&\color{orangered}{ 16 }&8\\& & 9& 3& \color{orangered}{1} & \\ \hline &9&3&1&\color{orangered}{17}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 17 } = \color{blue}{ 17 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&9&-6&-2&16&8\\& & 9& 3& 1& \color{blue}{17} \\ \hline &9&3&1&\color{blue}{17}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 17 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrrr}1&9&-6&-2&16&\color{orangered}{ 8 }\\& & 9& 3& 1& \color{orangered}{17} \\ \hline &\color{blue}{9}&\color{blue}{3}&\color{blue}{1}&\color{blue}{17}&\color{orangered}{25} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 25 }\right)$.