Divide using synthetic division $$ ( 4x^{4}-21x^{3}+7x^{2}-6x-14 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&4&-21&7&-6&-14\\& & 20& -5& 10& \color{black}{20} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 4x^{4}-21x^{3}+7x^{2}-6x-14 }{ x-5 } = \color{blue}{4x^{3}-x^{2}+2x+4} ~+~ \frac{ \color{red}{ 6 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-21&7&-6&-14\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 4 }&-21&7&-6&-14\\& & & & & \\ \hline &\color{orangered}{4}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-21&7&-6&-14\\& & \color{blue}{20} & & & \\ \hline &\color{blue}{4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 20 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&4&\color{orangered}{ -21 }&7&-6&-14\\& & \color{orangered}{20} & & & \\ \hline &4&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-21&7&-6&-14\\& & 20& \color{blue}{-5} & & \\ \hline &4&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}5&4&-21&\color{orangered}{ 7 }&-6&-14\\& & 20& \color{orangered}{-5} & & \\ \hline &4&-1&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 2 } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-21&7&-6&-14\\& & 20& -5& \color{blue}{10} & \\ \hline &4&-1&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 10 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&4&-21&7&\color{orangered}{ -6 }&-14\\& & 20& -5& \color{orangered}{10} & \\ \hline &4&-1&2&\color{orangered}{4}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&4&-21&7&-6&-14\\& & 20& -5& 10& \color{blue}{20} \\ \hline &4&-1&2&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 20 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}5&4&-21&7&-6&\color{orangered}{ -14 }\\& & 20& -5& 10& \color{orangered}{20} \\ \hline &\color{blue}{4}&\color{blue}{-1}&\color{blue}{2}&\color{blue}{4}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 4x^{3}-x^{2}+2x+4 } $ with a remainder of $ \color{red}{ 6 } $.