Find remainder when $ x^{4}+5x^{3}-8x+10 $ is divided by $ x+2 $.

The synthetic division table is:

$$ \begin{array}{c|rrrrr}-2&1&5&0&-8&10\\& & -2& -6& 12& \color{black}{-8} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{2} \end{array} $$

The remainder when $ x^{4}+5x^{3}-8x+10 $ is divided by $ x+2 $ 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 + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.

$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&5&0&-8&10\\& & & & & \\ \hline &&&&& \end{array} $$

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

$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&5&0&-8&10\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$

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

$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&5&0&-8&10\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$

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

$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 5 }&0&-8&10\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{3}&&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&5&0&-8&10\\& & -2& \color{blue}{-6} & & \\ \hline &1&\color{blue}{3}&&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $

$$ \begin{array}{c|rrrrr}-2&1&5&\color{orangered}{ 0 }&-8&10\\& & -2& \color{orangered}{-6} & & \\ \hline &1&3&\color{orangered}{-6}&& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&5&0&-8&10\\& & -2& -6& \color{blue}{12} & \\ \hline &1&3&\color{blue}{-6}&& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 12 } = \color{orangered}{ 4 } $

$$ \begin{array}{c|rrrrr}-2&1&5&0&\color{orangered}{ -8 }&10\\& & -2& -6& \color{orangered}{12} & \\ \hline &1&3&-6&\color{orangered}{4}& \end{array} $$

Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.

$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&5&0&-8&10\\& & -2& -6& 12& \color{blue}{-8} \\ \hline &1&3&-6&\color{blue}{4}& \end{array} $$

Step 9 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 2 } $

$$ \begin{array}{c|rrrrr}-2&1&5&0&-8&\color{orangered}{ 10 }\\& & -2& -6& 12& \color{orangered}{-8} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-6}&\color{blue}{4}&\color{orangered}{2} \end{array} $$

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

Synthetic Division Calculator