Check if $ 3x-2 $ is a factor of $ 9x^{4}-6x^{3}-2x^{2}+16x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&9&-6&-2&16&8\\& & 18& 24& 44& \color{black}{120} \\ \hline &\color{blue}{9}&\color{blue}{12}&\color{blue}{22}&\color{blue}{60}&\color{orangered}{128} \end{array} $$Because the remainder $ \left( \color{red}{ 128 } \right) $ is not zero, we conclude that the $ x-2 $ 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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&9&-6&-2&16&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 9 }&-6&-2&16&8\\& & & & & \\ \hline &\color{orangered}{9}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&9&-6&-2&16&8\\& & \color{blue}{18} & & & \\ \hline &\color{blue}{9}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 18 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&9&\color{orangered}{ -6 }&-2&16&8\\& & \color{orangered}{18} & & & \\ \hline &9&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&9&-6&-2&16&8\\& & 18& \color{blue}{24} & & \\ \hline &9&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 24 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrrr}2&9&-6&\color{orangered}{ -2 }&16&8\\& & 18& \color{orangered}{24} & & \\ \hline &9&12&\color{orangered}{22}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 22 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&9&-6&-2&16&8\\& & 18& 24& \color{blue}{44} & \\ \hline &9&12&\color{blue}{22}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 44 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrrr}2&9&-6&-2&\color{orangered}{ 16 }&8\\& & 18& 24& \color{orangered}{44} & \\ \hline &9&12&22&\color{orangered}{60}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 60 } = \color{blue}{ 120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&9&-6&-2&16&8\\& & 18& 24& 44& \color{blue}{120} \\ \hline &9&12&22&\color{blue}{60}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 120 } = \color{orangered}{ 128 } $
$$ \begin{array}{c|rrrrr}2&9&-6&-2&16&\color{orangered}{ 8 }\\& & 18& 24& 44& \color{orangered}{120} \\ \hline &\color{blue}{9}&\color{blue}{12}&\color{blue}{22}&\color{blue}{60}&\color{orangered}{128} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 128 }\right)$.