Find remainder when $ 2x^{4}-10x^{3}+4x^{2}+21x-20 $ is divided by $ x-4 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&2&-10&4&21&-20\\& & 8& -8& -16& \color{black}{20} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$The remainder when $ 2x^{4}-10x^{3}+4x^{2}+21x-20 $ is divided by $ x-4 $ 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 -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-10&4&21&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 2 }&-10&4&21&-20\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 2 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-10&4&21&-20\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -10 } + \color{orangered}{ 8 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}4&2&\color{orangered}{ -10 }&4&21&-20\\& & \color{orangered}{8} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-10&4&21&-20\\& & 8& \color{blue}{-8} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}4&2&-10&\color{orangered}{ 4 }&21&-20\\& & 8& \color{orangered}{-8} & & \\ \hline &2&-2&\color{orangered}{-4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-10&4&21&-20\\& & 8& -8& \color{blue}{-16} & \\ \hline &2&-2&\color{blue}{-4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -16 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}4&2&-10&4&\color{orangered}{ 21 }&-20\\& & 8& -8& \color{orangered}{-16} & \\ \hline &2&-2&-4&\color{orangered}{5}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 5 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&2&-10&4&21&-20\\& & 8& -8& -16& \color{blue}{20} \\ \hline &2&-2&-4&\color{blue}{5}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}4&2&-10&4&21&\color{orangered}{ -20 }\\& & 8& -8& -16& \color{orangered}{20} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-4}&\color{blue}{5}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.