Divide using synthetic division $$ ( 21x^{4}-39x^{3}-16x^{2}+28x+40 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&21&-39&-16&28&40\\& & 42& 6& -20& \color{black}{16} \\ \hline &\color{blue}{21}&\color{blue}{3}&\color{blue}{-10}&\color{blue}{8}&\color{orangered}{56} \end{array} $$The solution is:
$$ \frac{ 21x^{4}-39x^{3}-16x^{2}+28x+40 }{ x-2 } = \color{blue}{21x^{3}+3x^{2}-10x+8} ~+~ \frac{ \color{red}{ 56 } }{ x-2 } $$Explanation
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.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&21&-39&-16&28&40\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 21 }&-39&-16&28&40\\& & & & & \\ \hline &\color{orangered}{21}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 21 } = \color{blue}{ 42 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&21&-39&-16&28&40\\& & \color{blue}{42} & & & \\ \hline &\color{blue}{21}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -39 } + \color{orangered}{ 42 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrrr}2&21&\color{orangered}{ -39 }&-16&28&40\\& & \color{orangered}{42} & & & \\ \hline &21&\color{orangered}{3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&21&-39&-16&28&40\\& & 42& \color{blue}{6} & & \\ \hline &21&\color{blue}{3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -16 } + \color{orangered}{ 6 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}2&21&-39&\color{orangered}{ -16 }&28&40\\& & 42& \color{orangered}{6} & & \\ \hline &21&3&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&21&-39&-16&28&40\\& & 42& 6& \color{blue}{-20} & \\ \hline &21&3&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}2&21&-39&-16&\color{orangered}{ 28 }&40\\& & 42& 6& \color{orangered}{-20} & \\ \hline &21&3&-10&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 8 } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&21&-39&-16&28&40\\& & 42& 6& -20& \color{blue}{16} \\ \hline &21&3&-10&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 40 } + \color{orangered}{ 16 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrrr}2&21&-39&-16&28&\color{orangered}{ 40 }\\& & 42& 6& -20& \color{orangered}{16} \\ \hline &\color{blue}{21}&\color{blue}{3}&\color{blue}{-10}&\color{blue}{8}&\color{orangered}{56} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 21x^{3}+3x^{2}-10x+8 } $ with a remainder of $ \color{red}{ 56 } $.