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

The synthetic division table is:

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

The remainder when $ x^{3}+3x-4 $ is divided by $ x-1 $ is $ \, \color{red}{ 0 } $.

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.

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

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

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

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

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

Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 1 } = \color{orangered}{ 1 } $

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

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

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

Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 1 } = \color{orangered}{ 4 } $

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

Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.

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

Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 4 } = \color{orangered}{ 0 } $

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

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

Synthetic Division Calculator