Find remainder when $ x^{4}-4x^{3}-2x+4 $ is divided by $ x-1 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&-4&0&-2&4\\& & 1& -3& -3& \color{black}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{-1} \end{array} $$The remainder when $ x^{4}-4x^{3}-2x+4 $ is divided by $ x-1 $ is $ \, \color{red}{ -1 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&0&-2&4\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&-4&0&-2&4\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&0&-2&4\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 1 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ -4 }&0&-2&4\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&0&-2&4\\& & 1& \color{blue}{-3} & & \\ \hline &1&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}1&1&-4&\color{orangered}{ 0 }&-2&4\\& & 1& \color{orangered}{-3} & & \\ \hline &1&-3&\color{orangered}{-3}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&0&-2&4\\& & 1& -3& \color{blue}{-3} & \\ \hline &1&-3&\color{blue}{-3}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}1&1&-4&0&\color{orangered}{ -2 }&4\\& & 1& -3& \color{orangered}{-3} & \\ \hline &1&-3&-3&\color{orangered}{-5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&-4&0&-2&4\\& & 1& -3& -3& \color{blue}{-5} \\ \hline &1&-3&-3&\color{blue}{-5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}1&1&-4&0&-2&\color{orangered}{ 4 }\\& & 1& -3& -3& \color{orangered}{-5} \\ \hline &\color{blue}{1}&\color{blue}{-3}&\color{blue}{-3}&\color{blue}{-5}&\color{orangered}{-1} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -1 }\right) $.