Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&1&7&12&28&16\\& & 1& 8& 20& \color{black}{48} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{20}&\color{blue}{48}&\color{orangered}{64} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x-1 } = \color{blue}{x^{3}+8x^{2}+20x+48} ~+~ \frac{ \color{red}{ 64 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&7&12&28&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 1 }&7&12&28&16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&7&12&28&16\\& & \color{blue}{1} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 1 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}1&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{1} & & & \\ \hline &1&\color{orangered}{8}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&7&12&28&16\\& & 1& \color{blue}{8} & & \\ \hline &1&\color{blue}{8}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 8 } = \color{orangered}{ 20 } $
$$ \begin{array}{c|rrrrr}1&1&7&\color{orangered}{ 12 }&28&16\\& & 1& \color{orangered}{8} & & \\ \hline &1&8&\color{orangered}{20}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 20 } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&7&12&28&16\\& & 1& 8& \color{blue}{20} & \\ \hline &1&8&\color{blue}{20}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 20 } = \color{orangered}{ 48 } $
$$ \begin{array}{c|rrrrr}1&1&7&12&\color{orangered}{ 28 }&16\\& & 1& 8& \color{orangered}{20} & \\ \hline &1&8&20&\color{orangered}{48}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 48 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&1&7&12&28&16\\& & 1& 8& 20& \color{blue}{48} \\ \hline &1&8&20&\color{blue}{48}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 48 } = \color{orangered}{ 64 } $
$$ \begin{array}{c|rrrrr}1&1&7&12&28&\color{orangered}{ 16 }\\& & 1& 8& 20& \color{orangered}{48} \\ \hline &\color{blue}{1}&\color{blue}{8}&\color{blue}{20}&\color{blue}{48}&\color{orangered}{64} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+8x^{2}+20x+48 } $ with a remainder of $ \color{red}{ 64 } $.