Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x+8 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-8&1&7&12&28&16\\& & -8& 8& -160& \color{black}{1056} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{20}&\color{blue}{-132}&\color{orangered}{1072} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x+8 } = \color{blue}{x^{3}-x^{2}+20x-132} ~+~ \frac{ \color{red}{ 1072 } }{ x+8 } $$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}{ \left( -8 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-8&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{-8} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&7&12&28&16\\& & -8& \color{blue}{8} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 8 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}-8&1&7&\color{orangered}{ 12 }&28&16\\& & -8& \color{orangered}{8} & & \\ \hline &1&-1&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ 20 } = \color{blue}{ -160 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&7&12&28&16\\& & -8& 8& \color{blue}{-160} & \\ \hline &1&-1&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -160 \right) } = \color{orangered}{ -132 } $
$$ \begin{array}{c|rrrrr}-8&1&7&12&\color{orangered}{ 28 }&16\\& & -8& 8& \color{orangered}{-160} & \\ \hline &1&-1&20&\color{orangered}{-132}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -8 } \cdot \color{blue}{ \left( -132 \right) } = \color{blue}{ 1056 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-8}&1&7&12&28&16\\& & -8& 8& -160& \color{blue}{1056} \\ \hline &1&-1&20&\color{blue}{-132}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 1056 } = \color{orangered}{ 1072 } $
$$ \begin{array}{c|rrrrr}-8&1&7&12&28&\color{orangered}{ 16 }\\& & -8& 8& -160& \color{orangered}{1056} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{20}&\color{blue}{-132}&\color{orangered}{1072} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+20x-132 } $ with a remainder of $ \color{red}{ 1072 } $.