Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&1&7&12&28&16\\& & -2& -10& -4& \color{black}{-48} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{2}&\color{blue}{24}&\color{orangered}{-32} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x+2 } = \color{blue}{x^{3}+5x^{2}+2x+24} \color{red}{~-~} \frac{ \color{red}{ 32 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&12&28&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 1 }&7&12&28&16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 1 } = \color{blue}{ -2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&12&28&16\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-2&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{-2} & & & \\ \hline &1&\color{orangered}{5}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 5 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&12&28&16\\& & -2& \color{blue}{-10} & & \\ \hline &1&\color{blue}{5}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}-2&1&7&\color{orangered}{ 12 }&28&16\\& & -2& \color{orangered}{-10} & & \\ \hline &1&5&\color{orangered}{2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&12&28&16\\& & -2& -10& \color{blue}{-4} & \\ \hline &1&5&\color{blue}{2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrrr}-2&1&7&12&\color{orangered}{ 28 }&16\\& & -2& -10& \color{orangered}{-4} & \\ \hline &1&5&2&\color{orangered}{24}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 24 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&1&7&12&28&16\\& & -2& -10& -4& \color{blue}{-48} \\ \hline &1&5&2&\color{blue}{24}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -32 } $
$$ \begin{array}{c|rrrrr}-2&1&7&12&28&\color{orangered}{ 16 }\\& & -2& -10& -4& \color{orangered}{-48} \\ \hline &\color{blue}{1}&\color{blue}{5}&\color{blue}{2}&\color{blue}{24}&\color{orangered}{-32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+5x^{2}+2x+24 } $ with a remainder of $ \color{red}{ -32 } $.