Divide using synthetic division $$ ( x^{3}+2x^{2}-41x-42 ) \div ( x-7 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}7&1&2&-41&-42\\& & 7& 63& \color{black}{154} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{22}&\color{orangered}{112} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-41x-42 }{ x-7 } = \color{blue}{x^{2}+9x+22} ~+~ \frac{ \color{red}{ 112 } }{ x-7 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-41&-42\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}7&\color{orangered}{ 1 }&2&-41&-42\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 1 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-41&-42\\& & \color{blue}{7} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 7 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}7&1&\color{orangered}{ 2 }&-41&-42\\& & \color{orangered}{7} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-41&-42\\& & 7& \color{blue}{63} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 63 } = \color{orangered}{ 22 } $
$$ \begin{array}{c|rrrr}7&1&2&\color{orangered}{ -41 }&-42\\& & 7& \color{orangered}{63} & \\ \hline &1&9&\color{orangered}{22}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 7 } \cdot \color{blue}{ 22 } = \color{blue}{ 154 } $.
$$ \begin{array}{c|rrrr}\color{blue}{7}&1&2&-41&-42\\& & 7& 63& \color{blue}{154} \\ \hline &1&9&\color{blue}{22}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ 154 } = \color{orangered}{ 112 } $
$$ \begin{array}{c|rrrr}7&1&2&-41&\color{orangered}{ -42 }\\& & 7& 63& \color{orangered}{154} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{22}&\color{orangered}{112} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x+22 } $ with a remainder of $ \color{red}{ 112 } $.