Divide using synthetic division $$ ( x^{5}+7x^{4}+19x^{3}+25x^{2}+16x+4 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}3&1&7&19&25&16&4\\& & 3& 30& 147& 516& \color{black}{1596} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{49}&\color{blue}{172}&\color{blue}{532}&\color{orangered}{1600} \end{array} $$The solution is:
$$ \frac{ x^{5}+7x^{4}+19x^{3}+25x^{2}+16x+4 }{ x-3 } = \color{blue}{x^{4}+10x^{3}+49x^{2}+172x+532} ~+~ \frac{ \color{red}{ 1600 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}3&\color{orangered}{ 1 }&7&19&25&16&4\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & \color{blue}{3} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrrr}3&1&\color{orangered}{ 7 }&19&25&16&4\\& & \color{orangered}{3} & & & & \\ \hline &1&\color{orangered}{10}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & 3& \color{blue}{30} & & & \\ \hline &1&\color{blue}{10}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ 30 } = \color{orangered}{ 49 } $
$$ \begin{array}{c|rrrrrr}3&1&7&\color{orangered}{ 19 }&25&16&4\\& & 3& \color{orangered}{30} & & & \\ \hline &1&10&\color{orangered}{49}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 49 } = \color{blue}{ 147 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & 3& 30& \color{blue}{147} & & \\ \hline &1&10&\color{blue}{49}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 25 } + \color{orangered}{ 147 } = \color{orangered}{ 172 } $
$$ \begin{array}{c|rrrrrr}3&1&7&19&\color{orangered}{ 25 }&16&4\\& & 3& 30& \color{orangered}{147} & & \\ \hline &1&10&49&\color{orangered}{172}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 172 } = \color{blue}{ 516 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & 3& 30& 147& \color{blue}{516} & \\ \hline &1&10&49&\color{blue}{172}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 516 } = \color{orangered}{ 532 } $
$$ \begin{array}{c|rrrrrr}3&1&7&19&25&\color{orangered}{ 16 }&4\\& & 3& 30& 147& \color{orangered}{516} & \\ \hline &1&10&49&172&\color{orangered}{532}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 532 } = \color{blue}{ 1596 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{3}&1&7&19&25&16&4\\& & 3& 30& 147& 516& \color{blue}{1596} \\ \hline &1&10&49&172&\color{blue}{532}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 1596 } = \color{orangered}{ 1600 } $
$$ \begin{array}{c|rrrrrr}3&1&7&19&25&16&\color{orangered}{ 4 }\\& & 3& 30& 147& 516& \color{orangered}{1596} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{49}&\color{blue}{172}&\color{blue}{532}&\color{orangered}{1600} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}+10x^{3}+49x^{2}+172x+532 } $ with a remainder of $ \color{red}{ 1600 } $.