Divide using synthetic division $$ ( x^{4}+7x^{3}+12x^{2}+28x+16 ) \div ( x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}4&1&7&12&28&16\\& & 4& 44& 224& \color{black}{1008} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{56}&\color{blue}{252}&\color{orangered}{1024} \end{array} $$The solution is:
$$ \frac{ x^{4}+7x^{3}+12x^{2}+28x+16 }{ x-4 } = \color{blue}{x^{3}+11x^{2}+56x+252} ~+~ \frac{ \color{red}{ 1024 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&7&12&28&16\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}4&\color{orangered}{ 1 }&7&12&28&16\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 1 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&7&12&28&16\\& & \color{blue}{4} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 4 } = \color{orangered}{ 11 } $
$$ \begin{array}{c|rrrrr}4&1&\color{orangered}{ 7 }&12&28&16\\& & \color{orangered}{4} & & & \\ \hline &1&\color{orangered}{11}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 11 } = \color{blue}{ 44 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&7&12&28&16\\& & 4& \color{blue}{44} & & \\ \hline &1&\color{blue}{11}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 44 } = \color{orangered}{ 56 } $
$$ \begin{array}{c|rrrrr}4&1&7&\color{orangered}{ 12 }&28&16\\& & 4& \color{orangered}{44} & & \\ \hline &1&11&\color{orangered}{56}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 56 } = \color{blue}{ 224 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&7&12&28&16\\& & 4& 44& \color{blue}{224} & \\ \hline &1&11&\color{blue}{56}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 224 } = \color{orangered}{ 252 } $
$$ \begin{array}{c|rrrrr}4&1&7&12&\color{orangered}{ 28 }&16\\& & 4& 44& \color{orangered}{224} & \\ \hline &1&11&56&\color{orangered}{252}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 252 } = \color{blue}{ 1008 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{4}&1&7&12&28&16\\& & 4& 44& 224& \color{blue}{1008} \\ \hline &1&11&56&\color{blue}{252}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ 1008 } = \color{orangered}{ 1024 } $
$$ \begin{array}{c|rrrrr}4&1&7&12&28&\color{orangered}{ 16 }\\& & 4& 44& 224& \color{orangered}{1008} \\ \hline &\color{blue}{1}&\color{blue}{11}&\color{blue}{56}&\color{blue}{252}&\color{orangered}{1024} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+11x^{2}+56x+252 } $ with a remainder of $ \color{red}{ 1024 } $.