Find remainder when $ 4x^{4}+2x^{3}+5x+1 $ is divided by $ x+1 $.

The synthetic division table is:

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

The remainder when $ 4x^{4}+2x^{3}+5x+1 $ is divided by $ x+1 $ is $ \, \color{red}{ -2 } $.

We can find remainder using synthetic division method.

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.

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

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

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

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 4 } = \color{blue}{ -4 } $.

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

Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -2 } $

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

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 2 } $.

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

Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 2 } = \color{orangered}{ 2 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 2 } = \color{blue}{ -2 } $.

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

Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 3 } $

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

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 } $.

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

Step 9 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -2 } $

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

Remainder is the last entry in the bottom row $ \left(\color{red}{ -2 }\right) $.

Synthetic Division Calculator