Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&1&7&12&28&16\\& & 3& 30& 126& \color{black}{462} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{42}&\color{blue}{154}&\color{orangered}{478} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x-3 } = \color{blue}{x^{3}+10x^{2}+42x+154} ~+~ \frac{ \color{red}{ 478 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&7&12&28&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ 1 }&7&12&28&16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&7&12&28&16\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}3&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{3} & & & \\ \hline &1&\color{orangered}{10}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&7&12&28&16\\& & 3& \color{blue}{30} & & \\ \hline &1&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 30 } = \color{orangered}{ 42 } $
$$ \begin{array}{c|rrrrr}3&1&7&\color{orangered}{ 12 }&28&16\\& & 3& \color{orangered}{30} & & \\ \hline &1&10&\color{orangered}{42}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 42 } = \color{blue}{ 126 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&7&12&28&16\\& & 3& 30& \color{blue}{126} & \\ \hline &1&10&\color{blue}{42}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 126 } = \color{orangered}{ 154 } $
$$ \begin{array}{c|rrrrr}3&1&7&12&\color{orangered}{ 28 }&16\\& & 3& 30& \color{orangered}{126} & \\ \hline &1&10&42&\color{orangered}{154}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 154 } = \color{blue}{ 462 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&1&7&12&28&16\\& & 3& 30& 126& \color{blue}{462} \\ \hline &1&10&42&\color{blue}{154}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 462 } = \color{orangered}{ 478 } $
$$ \begin{array}{c|rrrrr}3&1&7&12&28&\color{orangered}{ 16 }\\& & 3& 30& 126& \color{orangered}{462} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{42}&\color{blue}{154}&\color{orangered}{478} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+10x^{2}+42x+154 } $ with a remainder of $ \color{red}{ 478 } $.