Divide using synthetic division $$ ( 7x^{3}+4x^{2}+4x+3 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}8&7&4&4&3\\& & 56& 480& \color{black}{3872} \\ \hline &\color{blue}{7}&\color{blue}{60}&\color{blue}{484}&\color{orangered}{3875} \end{array} $$The solution is:
$$ \frac{ 7x^{3}+4x^{2}+4x+3 }{ x-8 } = \color{blue}{7x^{2}+60x+484} ~+~ \frac{ \color{red}{ 3875 } }{ x-8 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -8 = 0 $ ( $ x = \color{blue}{ 8 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{8}&7&4&4&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}8&\color{orangered}{ 7 }&4&4&3\\& & & & \\ \hline &\color{orangered}{7}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 7 } = \color{blue}{ 56 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&7&4&4&3\\& & \color{blue}{56} & & \\ \hline &\color{blue}{7}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 56 } = \color{orangered}{ 60 } $
$$ \begin{array}{c|rrrr}8&7&\color{orangered}{ 4 }&4&3\\& & \color{orangered}{56} & & \\ \hline &7&\color{orangered}{60}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 60 } = \color{blue}{ 480 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&7&4&4&3\\& & 56& \color{blue}{480} & \\ \hline &7&\color{blue}{60}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 480 } = \color{orangered}{ 484 } $
$$ \begin{array}{c|rrrr}8&7&4&\color{orangered}{ 4 }&3\\& & 56& \color{orangered}{480} & \\ \hline &7&60&\color{orangered}{484}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 484 } = \color{blue}{ 3872 } $.
$$ \begin{array}{c|rrrr}\color{blue}{8}&7&4&4&3\\& & 56& 480& \color{blue}{3872} \\ \hline &7&60&\color{blue}{484}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 3872 } = \color{orangered}{ 3875 } $
$$ \begin{array}{c|rrrr}8&7&4&4&\color{orangered}{ 3 }\\& & 56& 480& \color{orangered}{3872} \\ \hline &\color{blue}{7}&\color{blue}{60}&\color{blue}{484}&\color{orangered}{3875} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 7x^{2}+60x+484 } $ with a remainder of $ \color{red}{ 3875 } $.