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