Divide using synthetic division $$ ( x^{3}+7x^{2}+9x-9 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&7&9&-9\\& & 3& 30& \color{black}{117} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{39}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ x^{3}+7x^{2}+9x-9 }{ x-3 } = \color{blue}{x^{2}+10x+39} ~+~ \frac{ \color{red}{ 108 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&7&9&-9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&7&9&-9\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&7&9&-9\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 7 }&9&-9\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&7&9&-9\\& & 3& \color{blue}{30} & \\ \hline &1&\color{blue}{10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 30 } = \color{orangered}{ 39 } $
$$ \begin{array}{c|rrrr}3&1&7&\color{orangered}{ 9 }&-9\\& & 3& \color{orangered}{30} & \\ \hline &1&10&\color{orangered}{39}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 39 } = \color{blue}{ 117 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&7&9&-9\\& & 3& 30& \color{blue}{117} \\ \hline &1&10&\color{blue}{39}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 117 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrr}3&1&7&9&\color{orangered}{ -9 }\\& & 3& 30& \color{orangered}{117} \\ \hline &\color{blue}{1}&\color{blue}{10}&\color{blue}{39}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+10x+39 } $ with a remainder of $ \color{red}{ 108 } $.