Check if $ x+1 $ is a factor of $ 2x^{4}-3x^{3}-4x^{2}+3x+2 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}-1&2&-3&-4&3&2\\& & -2& 5& -1& \color{black}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$

Because the remainder equals zero, we conclude that the $ x+1 $ is a factor of the $ 2x^{4}-3x^{3}-4x^{2}+3x+2 $.

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.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-4&3&2\\& & & & & \\ \hline &&&&& \end{array} $$

Step 1 : Bring down the leading coefficient to the bottom row.

$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&-3&-4&3&2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-4&3&2\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -5 } $

$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ -3 }&-4&3&2\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-5}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 5 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-4&3&2\\& & -2& \color{blue}{5} & & \\ \hline &2&\color{blue}{-5}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 5 } = \color{orangered}{ 1 } $

$$ \begin{array}{c|rrrrr}-1&2&-3&\color{orangered}{ -4 }&3&2\\& & -2& \color{orangered}{5} & & \\ \hline &2&-5&\color{orangered}{1}&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-4&3&2\\& & -2& 5& \color{blue}{-1} & \\ \hline &2&-5&\color{blue}{1}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 2 } $

$$ \begin{array}{c|rrrrr}-1&2&-3&-4&\color{orangered}{ 3 }&2\\& & -2& 5& \color{orangered}{-1} & \\ \hline &2&-5&1&\color{orangered}{2}& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&-3&-4&3&2\\& & -2& 5& -1& \color{blue}{-2} \\ \hline &2&-5&1&\color{blue}{2}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 0 } $

$$ \begin{array}{c|rrrrr}-1&2&-3&-4&3&\color{orangered}{ 2 }\\& & -2& 5& -1& \color{orangered}{-2} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{1}&\color{blue}{2}&\color{orangered}{0} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right)$.

Synthetic Division Calculator