Find remainder when $ x^{4}+4x^{3}-25x^{2}+18x-25 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&1&4&-25&18&-25\\& & 2& 12& -26& \color{black}{-16} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-13}&\color{blue}{-8}&\color{orangered}{-41} \end{array} $$The remainder when $ x^{4}+4x^{3}-25x^{2}+18x-25 $ is divided by $ x-2 $ is $ \, \color{red}{ -41 } $.
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 -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&-25&18&-25\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 1 }&4&-25&18&-25\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 1 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&-25&18&-25\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 2 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}2&1&\color{orangered}{ 4 }&-25&18&-25\\& & \color{orangered}{2} & & & \\ \hline &1&\color{orangered}{6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 6 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&-25&18&-25\\& & 2& \color{blue}{12} & & \\ \hline &1&\color{blue}{6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 12 } = \color{orangered}{ -13 } $
$$ \begin{array}{c|rrrrr}2&1&4&\color{orangered}{ -25 }&18&-25\\& & 2& \color{orangered}{12} & & \\ \hline &1&6&\color{orangered}{-13}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -13 \right) } = \color{blue}{ -26 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&-25&18&-25\\& & 2& 12& \color{blue}{-26} & \\ \hline &1&6&\color{blue}{-13}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -26 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrr}2&1&4&-25&\color{orangered}{ 18 }&-25\\& & 2& 12& \color{orangered}{-26} & \\ \hline &1&6&-13&\color{orangered}{-8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&1&4&-25&18&-25\\& & 2& 12& -26& \color{blue}{-16} \\ \hline &1&6&-13&\color{blue}{-8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ -41 } $
$$ \begin{array}{c|rrrrr}2&1&4&-25&18&\color{orangered}{ -25 }\\& & 2& 12& -26& \color{orangered}{-16} \\ \hline &\color{blue}{1}&\color{blue}{6}&\color{blue}{-13}&\color{blue}{-8}&\color{orangered}{-41} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ -41 }\right) $.