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