Divide using synthetic division $$ ( x^{3}+2x^{2}-41x-42 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&1&2&-41&-42\\& & 1& 3& \color{black}{-38} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-38}&\color{orangered}{-80} \end{array} $$The solution is:
$$ \frac{ x^{3}+2x^{2}-41x-42 }{ x-1 } = \color{blue}{x^{2}+3x-38} \color{red}{~-~} \frac{ \color{red}{ 80 } }{ x-1 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -1 = 0 $ ( $ x = \color{blue}{ 1 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-41&-42\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 1 }&2&-41&-42\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 1 } = \color{blue}{ 1 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-41&-42\\& & \color{blue}{1} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 1 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}1&1&\color{orangered}{ 2 }&-41&-42\\& & \color{orangered}{1} & & \\ \hline &1&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 3 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-41&-42\\& & 1& \color{blue}{3} & \\ \hline &1&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -41 } + \color{orangered}{ 3 } = \color{orangered}{ -38 } $
$$ \begin{array}{c|rrrr}1&1&2&\color{orangered}{ -41 }&-42\\& & 1& \color{orangered}{3} & \\ \hline &1&3&\color{orangered}{-38}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -38 \right) } = \color{blue}{ -38 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&1&2&-41&-42\\& & 1& 3& \color{blue}{-38} \\ \hline &1&3&\color{blue}{-38}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ \left( -38 \right) } = \color{orangered}{ -80 } $
$$ \begin{array}{c|rrrr}1&1&2&-41&\color{orangered}{ -42 }\\& & 1& 3& \color{orangered}{-38} \\ \hline &\color{blue}{1}&\color{blue}{3}&\color{blue}{-38}&\color{orangered}{-80} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+3x-38 } $ with a remainder of $ \color{red}{ -80 } $.