Find remainder when $ x^{3}+x+1 $ is divided by $ x-3 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}3&1&0&1&1\\& & 3& 9& \color{black}{30} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{31} \end{array} $$

The remainder when $ x^{3}+x+1 $ is divided by $ x-3 $ is $ \, \color{red}{ 31 } $.

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 -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.

$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&1&1\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&0&1&1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.

$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&1&1\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 3 } = \color{orangered}{ 3 } $

$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 0 }&1&1\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$

Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.

$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&1&1\\& & 3& \color{blue}{9} & \\ \hline &1&\color{blue}{3}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $

$$ \begin{array}{c|rrrr}3&1&0&\color{orangered}{ 1 }&1\\& & 3& \color{orangered}{9} & \\ \hline &1&3&\color{orangered}{10}& \end{array} $$

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.

$$ \begin{array}{c|rrrr}\color{blue}{3}&1&0&1&1\\& & 3& 9& \color{blue}{30} \\ \hline &1&3&\color{blue}{10}& \end{array} $$

Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 30 } = \color{orangered}{ 31 } $

$$ \begin{array}{c|rrrr}3&1&0&1&\color{orangered}{ 1 }\\& & 3& 9& \color{orangered}{30} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{31} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 31 }\right) $.

Synthetic Division Calculator