Find remainder when $ x^{3}+4x^{2}-6x+2 $ is divided by $ x-1 $.

The synthetic division table is:

$$ \begin{array}{c|rrrr}1&1&4&-6&2\\& & 1& 5& \color{black}{-1} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-1}&\color{orangered}{1} \end{array} $$

The remainder when $ x^{3}+4x^{2}-6x+2 $ is divided by $ x-1 $ is $ \, \color{red}{ 1 } $.

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.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-6&2\\& & & & \\ \hline &&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&4&-6&2\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$

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 } $.

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-6&2\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 4 }&-6&2\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{5}&& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-6&2\\& & 1& \color{blue}{5} & \\ \hline &1&\color{blue}{5}&& \end{array} $$

Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 5 } = \color{orangered}{ -1 } $

$$ \begin{array}{c|rrrr}1&1&4&\color{orangered}{ -6 }&2\\& & 1& \color{orangered}{5} & \\ \hline &1&5&\color{orangered}{-1}& \end{array} $$

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

$$ \begin{array}{c|rrrr}\color{blue}{1}&1&4&-6&2\\& & 1& 5& \color{blue}{-1} \\ \hline &1&5&\color{blue}{-1}& \end{array} $$

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

$$ \begin{array}{c|rrrr}1&1&4&-6&\color{orangered}{ 2 }\\& & 1& 5& \color{orangered}{-1} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{-1}&\color{orangered}{1} \end{array} $$

Remainder is the last entry in the bottom row $ \left(\color{red}{ 1 }\right) $.

Synthetic Division Calculator