Divide using synthetic division $$ ( x^{4}-5x^{3}+7x^{2}-34x-1 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&1&-5&7&-34&-1\\& & 5& 0& 35& \color{black}{5} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{7}&\color{blue}{1}&\color{orangered}{4} \end{array} $$The solution is:
$$ \frac{ x^{4}-5x^{3}+7x^{2}-34x-1 }{ x-5 } = \color{blue}{x^{3}+7x+1} ~+~ \frac{ \color{red}{ 4 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-5&7&-34&-1\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ 1 }&-5&7&-34&-1\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-5&7&-34&-1\\& & \color{blue}{5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 5 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&1&\color{orangered}{ -5 }&7&-34&-1\\& & \color{orangered}{5} & & & \\ \hline &1&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-5&7&-34&-1\\& & 5& \color{blue}{0} & & \\ \hline &1&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 0 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrr}5&1&-5&\color{orangered}{ 7 }&-34&-1\\& & 5& \color{orangered}{0} & & \\ \hline &1&0&\color{orangered}{7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 7 } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-5&7&-34&-1\\& & 5& 0& \color{blue}{35} & \\ \hline &1&0&\color{blue}{7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -34 } + \color{orangered}{ 35 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}5&1&-5&7&\color{orangered}{ -34 }&-1\\& & 5& 0& \color{orangered}{35} & \\ \hline &1&0&7&\color{orangered}{1}& \end{array} $$Step 8 : 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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&1&-5&7&-34&-1\\& & 5& 0& 35& \color{blue}{5} \\ \hline &1&0&7&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 5 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}5&1&-5&7&-34&\color{orangered}{ -1 }\\& & 5& 0& 35& \color{orangered}{5} \\ \hline &\color{blue}{1}&\color{blue}{0}&\color{blue}{7}&\color{blue}{1}&\color{orangered}{4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+7x+1 } $ with a remainder of $ \color{red}{ 4 } $.