Divide using synthetic division $$ ( x^{3}+2x^{2}-2x+9 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&2&-2&9\\& & 7& 63& \color{black}{427} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{61}&\color{orangered}{436} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-2x+9 }{ x-7 } = \color{blue}{x^{2}+9x+61} ~+~ \frac{ \color{red}{ 436 } }{ x-7 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -7 = 0 $ ( $ x = \color{blue}{ 7 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-2&9\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&2&-2&9\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-2&9\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 7 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ 2 }&-2&9\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 9 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-2&9\\& & 7& \color{blue}{63} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ 63 } = \color{orangered}{ 61 } $
$$ \begin{array}{c|rrrr}7&1&2&\color{orangered}{ -2 }&9\\& & 7& \color{orangered}{63} & \\ \hline &1&9&\color{orangered}{61}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 61 } = \color{blue}{ 427 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-2&9\\& & 7& 63& \color{blue}{427} \\ \hline &1&9&\color{blue}{61}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 427 } = \color{orangered}{ 436 } $
$$ \begin{array}{c|rrrr}7&1&2&-2&\color{orangered}{ 9 }\\& & 7& 63& \color{orangered}{427} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{61}&\color{orangered}{436} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x+61 } $ with a remainder of $ \color{red}{ 436 } $.