Divide using synthetic division $$ ( x^{3}-4x^{2}+5x-42 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&1&-4&5&-42\\& & 5& 5& \color{black}{50} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{10}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ x^{3}-4x^{2}+5x-42 }{ x-5 } = \color{blue}{x^{2}+x+10} ~+~ \frac{ \color{red}{ 8 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&5&-42\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 1 }&-4&5&-42\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&5&-42\\& & \color{blue}{5} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 5 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}5&1&\color{orangered}{ -4 }&5&-42\\& & \color{orangered}{5} & & \\ \hline &1&\color{orangered}{1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 1 } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&5&-42\\& & 5& \color{blue}{5} & \\ \hline &1&\color{blue}{1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 5 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}5&1&-4&\color{orangered}{ 5 }&-42\\& & 5& \color{orangered}{5} & \\ \hline &1&1&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 10 } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&1&-4&5&-42\\& & 5& 5& \color{blue}{50} \\ \hline &1&1&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -42 } + \color{orangered}{ 50 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrr}5&1&-4&5&\color{orangered}{ -42 }\\& & 5& 5& \color{orangered}{50} \\ \hline &\color{blue}{1}&\color{blue}{1}&\color{blue}{10}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+x+10 } $ with a remainder of $ \color{red}{ 8 } $.