Find remainder when $ 2x^{4}+3x^{3}-6x^{2}-9x-2 $ is divided by $ x-1 $.
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&3&-6&-9&-2\\& & 2& 5& -1& \color{black}{-10} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-1}&\color{blue}{-10}&\color{orangered}{-12} \end{array} $$The remainder when $ 2x^{4}+3x^{3}-6x^{2}-9x-2 $ is divided by $ x-1 $ is $ \, \color{red}{ -12 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&3&-6&-9&-2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 2 }&3&-6&-9&-2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&3&-6&-9&-2\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 2 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ 3 }&-6&-9&-2\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{5}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&3&-6&-9&-2\\& & 2& \color{blue}{5} & & \\ \hline &2&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 5 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}1&2&3&\color{orangered}{ -6 }&-9&-2\\& & 2& \color{orangered}{5} & & \\ \hline &2&5&\color{orangered}{-1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&3&-6&-9&-2\\& & 2& 5& \color{blue}{-1} & \\ \hline &2&5&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}1&2&3&-6&\color{orangered}{ -9 }&-2\\& & 2& 5& \color{orangered}{-1} & \\ \hline &2&5&-1&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&3&-6&-9&-2\\& & 2& 5& -1& \color{blue}{-10} \\ \hline &2&5&-1&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}1&2&3&-6&-9&\color{orangered}{ -2 }\\& & 2& 5& -1& \color{orangered}{-10} \\ \hline &\color{blue}{2}&\color{blue}{5}&\color{blue}{-1}&\color{blue}{-10}&\color{orangered}{-12} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -12 }\right) $.