Divide using synthetic division $$ ( 2x^{4}+7x^{3}+18 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}1&2&7&0&0&18\\& & 2& 9& 9& \color{black}{9} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{9}&\color{blue}{9}&\color{orangered}{27} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+7x^{3}+18 }{ x-1 } = \color{blue}{2x^{3}+9x^{2}+9x+9} ~+~ \frac{ \color{red}{ 27 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&0&0&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}1&\color{orangered}{ 2 }&7&0&0&18\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&0&0&18\\& & \color{blue}{2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 2 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&2&\color{orangered}{ 7 }&0&0&18\\& & \color{orangered}{2} & & & \\ \hline &2&\color{orangered}{9}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&0&0&18\\& & 2& \color{blue}{9} & & \\ \hline &2&\color{blue}{9}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&2&7&\color{orangered}{ 0 }&0&18\\& & 2& \color{orangered}{9} & & \\ \hline &2&9&\color{orangered}{9}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&0&0&18\\& & 2& 9& \color{blue}{9} & \\ \hline &2&9&\color{blue}{9}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 9 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}1&2&7&0&\color{orangered}{ 0 }&18\\& & 2& 9& \color{orangered}{9} & \\ \hline &2&9&9&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 9 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{1}&2&7&0&0&18\\& & 2& 9& 9& \color{blue}{9} \\ \hline &2&9&9&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ 9 } = \color{orangered}{ 27 } $
$$ \begin{array}{c|rrrrr}1&2&7&0&0&\color{orangered}{ 18 }\\& & 2& 9& 9& \color{orangered}{9} \\ \hline &\color{blue}{2}&\color{blue}{9}&\color{blue}{9}&\color{blue}{9}&\color{orangered}{27} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+9x^{2}+9x+9 } $ with a remainder of $ \color{red}{ 27 } $.