Find remainder when $ 4x^{5}+21x^{4}+2x^{3}-11x^{2}+20x $ is divided by $ x+5 $.
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&4&21&2&-11&20&0\\& & -20& -5& 15& -20& \color{black}{0} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The remainder when $ 4x^{5}+21x^{4}+2x^{3}-11x^{2}+20x $ is divided by $ x+5 $ 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 + 5 = 0 $ ( $ x = \color{blue}{ -5 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 4 }&21&2&-11&20&0\\& & & & & & \\ \hline &\color{orangered}{4}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & \color{blue}{-20} & & & & \\ \hline &\color{blue}{4}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 21 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-5&4&\color{orangered}{ 21 }&2&-11&20&0\\& & \color{orangered}{-20} & & & & \\ \hline &4&\color{orangered}{1}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & -20& \color{blue}{-5} & & & \\ \hline &4&\color{blue}{1}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrrr}-5&4&21&\color{orangered}{ 2 }&-11&20&0\\& & -20& \color{orangered}{-5} & & & \\ \hline &4&1&\color{orangered}{-3}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & -20& -5& \color{blue}{15} & & \\ \hline &4&1&\color{blue}{-3}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -11 } + \color{orangered}{ 15 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrrr}-5&4&21&2&\color{orangered}{ -11 }&20&0\\& & -20& -5& \color{orangered}{15} & & \\ \hline &4&1&-3&\color{orangered}{4}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & -20& -5& 15& \color{blue}{-20} & \\ \hline &4&1&-3&\color{blue}{4}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-5&4&21&2&-11&\color{orangered}{ 20 }&0\\& & -20& -5& 15& \color{orangered}{-20} & \\ \hline &4&1&-3&4&\color{orangered}{0}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&4&21&2&-11&20&0\\& & -20& -5& 15& -20& \color{blue}{0} \\ \hline &4&1&-3&4&\color{blue}{0}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-5&4&21&2&-11&20&\color{orangered}{ 0 }\\& & -20& -5& 15& -20& \color{orangered}{0} \\ \hline &\color{blue}{4}&\color{blue}{1}&\color{blue}{-3}&\color{blue}{4}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Remainder is the last entry in the bottom row $ \left(\color{red}{ 0 }\right) $.