Divide using synthetic division $$ ( 2x^{3}+2x^{2}+4x+14 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&2&4&14\\& & 2& 4& \color{black}{8} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{8}&\color{orangered}{22} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+2x^{2}+4x+14 }{ x-1 } = \color{blue}{2x^{2}+4x+8} ~+~ \frac{ \color{red}{ 22 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&2&4&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&2&4&14\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&2&4&14\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ 2 }&4&14\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 4 } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&2&4&14\\& & 2& \color{blue}{4} & \\ \hline &2&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 4 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}1&2&2&\color{orangered}{ 4 }&14\\& & 2& \color{orangered}{4} & \\ \hline &2&4&\color{orangered}{8}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 8 } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&2&4&14\\& & 2& 4& \color{blue}{8} \\ \hline &2&4&\color{blue}{8}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 8 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}1&2&2&4&\color{orangered}{ 14 }\\& & 2& 4& \color{orangered}{8} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{8}&\color{orangered}{22} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+4x+8 } $ with a remainder of $ \color{red}{ 22 } $.