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