Divide using synthetic division $$ ( x^{3}+2x^{2}-41x-42 ) \div ( x+9 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-9&1&2&-41&-42\\& & -9& 63& \color{black}{-198} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{22}&\color{orangered}{-240} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-41x-42 }{ x+9 } = \color{blue}{x^{2}-7x+22} \color{red}{~-~} \frac{ \color{red}{ 240 } }{ x+9 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-41&-42\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-9&\color{orangered}{ 1 }&2&-41&-42\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-41&-42\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}-9&1&\color{orangered}{ 2 }&-41&-42\\& & \color{orangered}{-9} & & \\ \hline &1&\color{orangered}{-7}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-41&-42\\& & -9& \color{blue}{63} & \\ \hline &1&\color{blue}{-7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 63 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}-9&1&2&\color{orangered}{ -41 }&-42\\& & -9& \color{orangered}{63} & \\ \hline &1&-7&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -9 } \cdot \color{blue}{ 22 } = \color{blue}{ -198 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-9}&1&2&-41&-42\\& & -9& 63& \color{blue}{-198} \\ \hline &1&-7&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ \left( -198 \right) } = \color{orangered}{ -240 } $
$$ \begin{array}{c|rrrr}-9&1&2&-41&\color{orangered}{ -42 }\\& & -9& 63& \color{orangered}{-198} \\ \hline &\color{blue}{1}&\color{blue}{-7}&\color{blue}{22}&\color{orangered}{-240} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}-7x+22 } $ with a remainder of $ \color{red}{ -240 } $.