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