Divide using synthetic division $$ ( 3x^{4}+9x^{3}-9x+9 ) \div ( x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}2&3&9&0&-9&9\\& & 6& 30& 60& \color{black}{102} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{30}&\color{blue}{51}&\color{orangered}{111} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+9x^{3}-9x+9 }{ x-2 } = \color{blue}{3x^{3}+15x^{2}+30x+51} ~+~ \frac{ \color{red}{ 111 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&9&0&-9&9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}2&\color{orangered}{ 3 }&9&0&-9&9\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 3 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&9&0&-9&9\\& & \color{blue}{6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 6 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}2&3&\color{orangered}{ 9 }&0&-9&9\\& & \color{orangered}{6} & & & \\ \hline &3&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 15 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&9&0&-9&9\\& & 6& \color{blue}{30} & & \\ \hline &3&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrrr}2&3&9&\color{orangered}{ 0 }&-9&9\\& & 6& \color{orangered}{30} & & \\ \hline &3&15&\color{orangered}{30}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 30 } = \color{blue}{ 60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&9&0&-9&9\\& & 6& 30& \color{blue}{60} & \\ \hline &3&15&\color{blue}{30}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 60 } = \color{orangered}{ 51 } $
$$ \begin{array}{c|rrrrr}2&3&9&0&\color{orangered}{ -9 }&9\\& & 6& 30& \color{orangered}{60} & \\ \hline &3&15&30&\color{orangered}{51}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 51 } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{2}&3&9&0&-9&9\\& & 6& 30& 60& \color{blue}{102} \\ \hline &3&15&30&\color{blue}{51}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 102 } = \color{orangered}{ 111 } $
$$ \begin{array}{c|rrrrr}2&3&9&0&-9&\color{orangered}{ 9 }\\& & 6& 30& 60& \color{orangered}{102} \\ \hline &\color{blue}{3}&\color{blue}{15}&\color{blue}{30}&\color{blue}{51}&\color{orangered}{111} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+15x^{2}+30x+51 } $ with a remainder of $ \color{red}{ 111 } $.