Divide using synthetic division $$ ( x^{4}-20x^{2}-21x-20 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&0&-20&-21&-20\\& & 5& 25& 25& \color{black}{20} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ x^{4}-20x^{2}-21x-20 }{ x-5 } = \color{blue}{x^{3}+5x^{2}+5x+4} $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-20&-21&-20\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&0&-20&-21&-20\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-20&-21&-20\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ 0 }&-20&-21&-20\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-20&-21&-20\\& & 5& \color{blue}{25} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 25 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}5&1&0&\color{orangered}{ -20 }&-21&-20\\& & 5& \color{orangered}{25} & & \\ \hline &1&5&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 5 } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-20&-21&-20\\& & 5& 25& \color{blue}{25} & \\ \hline &1&5&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 25 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&1&0&-20&\color{orangered}{ -21 }&-20\\& & 5& 25& \color{orangered}{25} & \\ \hline &1&5&5&\color{orangered}{4}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&0&-20&-21&-20\\& & 5& 25& 25& \color{blue}{20} \\ \hline &1&5&5&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -20 } + \color{orangered}{ 20 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&0&-20&-21&\color{orangered}{ -20 }\\& & 5& 25& 25& \color{orangered}{20} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{5}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+5x+4 } $ with a remainder of $ \color{red}{ 0 } $.