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