Find remainder when $ 5x^{4}-2x^{3}-4x^{2}+2 $ is divided by $ x-2 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&5&-2&-4&0&2\\& & 10& 16& 24& \color{black}{48} \\ \hline &\color{blue}{5}&\color{blue}{8}&\color{blue}{12}&\color{blue}{24}&\color{orangered}{50} \end{array} $$The remainder when $ 5x^{4}-2x^{3}-4x^{2}+2 $ is divided by $ x-2 $ is $ \, \color{red}{ 50 } $.
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-2&-4&0&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 5 }&-2&-4&0&2\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 5 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-2&-4&0&2\\& & \color{blue}{10} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 10 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&5&\color{orangered}{ -2 }&-4&0&2\\& & \color{orangered}{10} & & & \\ \hline &5&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-2&-4&0&2\\& & 10& \color{blue}{16} & & \\ \hline &5&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 16 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}2&5&-2&\color{orangered}{ -4 }&0&2\\& & 10& \color{orangered}{16} & & \\ \hline &5&8&\color{orangered}{12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 12 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-2&-4&0&2\\& & 10& 16& \color{blue}{24} & \\ \hline &5&8&\color{blue}{12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 24 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}2&5&-2&-4&\color{orangered}{ 0 }&2\\& & 10& 16& \color{orangered}{24} & \\ \hline &5&8&12&\color{orangered}{24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 24 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&5&-2&-4&0&2\\& & 10& 16& 24& \color{blue}{48} \\ \hline &5&8&12&\color{blue}{24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 48 } = \color{orangered}{ 50 } $
$$ \begin{array}{c|rrrrr}2&5&-2&-4&0&\color{orangered}{ 2 }\\& & 10& 16& 24& \color{orangered}{48} \\ \hline &\color{blue}{5}&\color{blue}{8}&\color{blue}{12}&\color{blue}{24}&\color{orangered}{50} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 50 }\right) $.