Check if $ x+1 $ is a factor of $ 9x^{4}-6x^{3}-2x^{2}+16x+8 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&9&-6&-2&16&8\\& & -9& 15& -13& \color{black}{-3} \\ \hline &\color{blue}{9}&\color{blue}{-15}&\color{blue}{13}&\color{blue}{3}&\color{orangered}{5} \end{array} $$Because the remainder $ \left( \color{red}{ 5 } \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}{ \left( -9 \right) } = \color{orangered}{ -15 } $
$$ \begin{array}{c|rrrrr}-1&9&\color{orangered}{ -6 }&-2&16&8\\& & \color{orangered}{-9} & & & \\ \hline &9&\color{orangered}{-15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -15 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&9&-6&-2&16&8\\& & -9& \color{blue}{15} & & \\ \hline &9&\color{blue}{-15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 15 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrrr}-1&9&-6&\color{orangered}{ -2 }&16&8\\& & -9& \color{orangered}{15} & & \\ \hline &9&-15&\color{orangered}{13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 13 } = \color{blue}{ -13 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&9&-6&-2&16&8\\& & -9& 15& \color{blue}{-13} & \\ \hline &9&-15&\color{blue}{13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -13 \right) } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}-1&9&-6&-2&\color{orangered}{ 16 }&8\\& & -9& 15& \color{orangered}{-13} & \\ \hline &9&-15&13&\color{orangered}{3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 3 } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&9&-6&-2&16&8\\& & -9& 15& -13& \color{blue}{-3} \\ \hline &9&-15&13&\color{blue}{3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-1&9&-6&-2&16&\color{orangered}{ 8 }\\& & -9& 15& -13& \color{orangered}{-3} \\ \hline &\color{blue}{9}&\color{blue}{-15}&\color{blue}{13}&\color{blue}{3}&\color{orangered}{5} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 5 }\right)$.