Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x-8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}8&1&7&12&28&16\\& & 8& 120& 1056& \color{black}{8672} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{132}&\color{blue}{1084}&\color{orangered}{8688} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x-8 } = \color{blue}{x^{3}+15x^{2}+132x+1084} ~+~ \frac{ \color{red}{ 8688 } }{ x-8 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&7&12&28&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}8&\color{orangered}{ 1 }&7&12&28&16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&7&12&28&16\\& & \color{blue}{8} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 8 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}8&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{8} & & & \\ \hline &1&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 15 } = \color{blue}{ 120 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&7&12&28&16\\& & 8& \color{blue}{120} & & \\ \hline &1&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 120 } = \color{orangered}{ 132 } $
$$ \begin{array}{c|rrrrr}8&1&7&\color{orangered}{ 12 }&28&16\\& & 8& \color{orangered}{120} & & \\ \hline &1&15&\color{orangered}{132}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 132 } = \color{blue}{ 1056 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&7&12&28&16\\& & 8& 120& \color{blue}{1056} & \\ \hline &1&15&\color{blue}{132}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 1056 } = \color{orangered}{ 1084 } $
$$ \begin{array}{c|rrrrr}8&1&7&12&\color{orangered}{ 28 }&16\\& & 8& 120& \color{orangered}{1056} & \\ \hline &1&15&132&\color{orangered}{1084}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 8 } \cdot \color{blue}{ 1084 } = \color{blue}{ 8672 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{8}&1&7&12&28&16\\& & 8& 120& 1056& \color{blue}{8672} \\ \hline &1&15&132&\color{blue}{1084}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 8672 } = \color{orangered}{ 8688 } $
$$ \begin{array}{c|rrrrr}8&1&7&12&28&\color{orangered}{ 16 }\\& & 8& 120& 1056& \color{orangered}{8672} \\ \hline &\color{blue}{1}&\color{blue}{15}&\color{blue}{132}&\color{blue}{1084}&\color{orangered}{8688} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+15x^{2}+132x+1084 } $ with a remainder of $ \color{red}{ 8688 } $.