Check if $ 3x+2 $ is a factor of $ 9x^{4}-6x^{3}-2x^{2}+16x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&9&-6&-2&16&8\\& & -18& 48& -92& \color{black}{152} \\ \hline &\color{blue}{9}&\color{blue}{-24}&\color{blue}{46}&\color{blue}{-76}&\color{orangered}{160} \end{array} $$Because the remainder $ \left( \color{red}{ 160 } \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}{ \left( -18 \right) } = \color{orangered}{ -24 } $
$$ \begin{array}{c|rrrrr}-2&9&\color{orangered}{ -6 }&-2&16&8\\& & \color{orangered}{-18} & & & \\ \hline &9&\color{orangered}{-24}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -24 \right) } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&9&-6&-2&16&8\\& & -18& \color{blue}{48} & & \\ \hline &9&\color{blue}{-24}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 48 } = \color{orangered}{ 46 } $
$$ \begin{array}{c|rrrrr}-2&9&-6&\color{orangered}{ -2 }&16&8\\& & -18& \color{orangered}{48} & & \\ \hline &9&-24&\color{orangered}{46}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 46 } = \color{blue}{ -92 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&9&-6&-2&16&8\\& & -18& 48& \color{blue}{-92} & \\ \hline &9&-24&\color{blue}{46}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -92 \right) } = \color{orangered}{ -76 } $
$$ \begin{array}{c|rrrrr}-2&9&-6&-2&\color{orangered}{ 16 }&8\\& & -18& 48& \color{orangered}{-92} & \\ \hline &9&-24&46&\color{orangered}{-76}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -76 \right) } = \color{blue}{ 152 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&9&-6&-2&16&8\\& & -18& 48& -92& \color{blue}{152} \\ \hline &9&-24&46&\color{blue}{-76}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 152 } = \color{orangered}{ 160 } $
$$ \begin{array}{c|rrrrr}-2&9&-6&-2&16&\color{orangered}{ 8 }\\& & -18& 48& -92& \color{orangered}{152} \\ \hline &\color{blue}{9}&\color{blue}{-24}&\color{blue}{46}&\color{blue}{-76}&\color{orangered}{160} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 160 }\right)$.