Check if $ x+2 $ is a factor of $ 2x^{3}-5x^{2}+x+2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-5&1&2\\& & -4& 18& \color{black}{-38} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{19}&\color{orangered}{-36} \end{array} $$Because the remainder $ \left( \color{red}{ -36 } \right) $ is not zero, we conclude that the $ x+2 $ is not a factor of $ 2x^{3}-5x^{2}+x+2$.
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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&1&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-5&1&2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&1&2\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -5 }&1&2\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&1&2\\& & -4& \color{blue}{18} & \\ \hline &2&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 18 } = \color{orangered}{ 19 } $
$$ \begin{array}{c|rrrr}-2&2&-5&\color{orangered}{ 1 }&2\\& & -4& \color{orangered}{18} & \\ \hline &2&-9&\color{orangered}{19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 19 } = \color{blue}{ -38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&1&2\\& & -4& 18& \color{blue}{-38} \\ \hline &2&-9&\color{blue}{19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -38 \right) } = \color{orangered}{ -36 } $
$$ \begin{array}{c|rrrr}-2&2&-5&1&\color{orangered}{ 2 }\\& & -4& 18& \color{orangered}{-38} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{19}&\color{orangered}{-36} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -36 }\right)$.