Check if $ x+2 $ is a factor of $ 2x^{3}-x^{2}-2x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-1&-2&-1\\& & -4& 10& \color{black}{-16} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{8}&\color{orangered}{-17} \end{array} $$Because the remainder $ \left( \color{red}{ -17 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ 2x^{3}-x^{2}-2x-1$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&-2&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-1&-2&-1\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&-2&-1\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -1 }&-2&-1\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&-2&-1\\& & -4& \color{blue}{10} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 10 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}-2&2&-1&\color{orangered}{ -2 }&-1\\& & -4& \color{orangered}{10} & \\ \hline &2&-5&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 8 } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&-2&-1\\& & -4& 10& \color{blue}{-16} \\ \hline &2&-5&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}-2&2&-1&-2&\color{orangered}{ -1 }\\& & -4& 10& \color{orangered}{-16} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{8}&\color{orangered}{-17} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -17 }\right)$.