Find remainder when $ x^{3}+2x^{2}-7x+1 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&2&-7&1\\& & 1& 3& \color{black}{-4} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-4}&\color{orangered}{-3} \end{array} $$The remainder when $ x^{3}+2x^{2}-7x+1 $ is divided by $ x-1 $ is $ \, \color{red}{ -3 } $.
Explanation
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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-7&1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&2&-7&1\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-7&1\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 2 }&-7&1\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-7&1\\& & 1& \color{blue}{3} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 3 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}1&1&2&\color{orangered}{ -7 }&1\\& & 1& \color{orangered}{3} & \\ \hline &1&3&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-7&1\\& & 1& 3& \color{blue}{-4} \\ \hline &1&3&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}1&1&2&-7&\color{orangered}{ 1 }\\& & 1& 3& \color{orangered}{-4} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-4}&\color{orangered}{-3} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -3 }\right) $.