Divide using synthetic division $$ ( x^{3}+2x^{2}-53x+89 ) \div ( x+9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-9&1&2&-53&89\\& & -9& 63& \color{black}{-90} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{10}&\color{orangered}{-1} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-53x+89 }{ x+9 } = \color{blue}{x^{2}-7x+10} \color{red}{~-~} \frac{ \color{red}{ 1 } }{ x+9 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 9 = 0 $ ( $ x = \color{blue}{ -9 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-53&89\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-9&\color{orangered}{ 1 }&2&-53&89\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 1 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-53&89\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-9&1&\color{orangered}{ 2 }&-53&89\\& & \color{orangered}{-9} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-53&89\\& & -9& \color{blue}{63} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -53 } + \color{orangered}{ 63 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}-9&1&2&\color{orangered}{ -53 }&89\\& & -9& \color{orangered}{63} & \\ \hline &1&-7&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 10 } = \color{blue}{ -90 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-53&89\\& & -9& 63& \color{blue}{-90} \\ \hline &1&-7&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 89 } + \color{orangered}{ \left( -90 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-9&1&2&-53&\color{orangered}{ 89 }\\& & -9& 63& \color{orangered}{-90} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{10}&\color{orangered}{-1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x+10 } $ with a remainder of $ \color{red}{ -1 } $.