Divide using synthetic division $$ ( x^{4}+4x^{3}+16x-35 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&1&4&0&16&-35\\& & -5& 5& -25& \color{black}{45} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{5}&\color{blue}{-9}&\color{orangered}{10} \end{array} $$The solution is:
$$ \frac{ x^{4}+4x^{3}+16x-35 }{ x+5 } = \color{blue}{x^{3}-x^{2}+5x-9} ~+~ \frac{ \color{red}{ 10 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&4&0&16&-35\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 1 }&4&0&16&-35\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&4&0&16&-35\\& & \color{blue}{-5} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-5&1&\color{orangered}{ 4 }&0&16&-35\\& & \color{orangered}{-5} & & & \\ \hline &1&\color{orangered}{-1}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 5 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&4&0&16&-35\\& & -5& \color{blue}{5} & & \\ \hline &1&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 5 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrrr}-5&1&4&\color{orangered}{ 0 }&16&-35\\& & -5& \color{orangered}{5} & & \\ \hline &1&-1&\color{orangered}{5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 5 } = \color{blue}{ -25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&4&0&16&-35\\& & -5& 5& \color{blue}{-25} & \\ \hline &1&-1&\color{blue}{5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 16 } + \color{orangered}{ \left( -25 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-5&1&4&0&\color{orangered}{ 16 }&-35\\& & -5& 5& \color{orangered}{-25} & \\ \hline &1&-1&5&\color{orangered}{-9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&1&4&0&16&-35\\& & -5& 5& -25& \color{blue}{45} \\ \hline &1&-1&5&\color{blue}{-9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -35 } + \color{orangered}{ 45 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-5&1&4&0&16&\color{orangered}{ -35 }\\& & -5& 5& -25& \color{orangered}{45} \\ \hline &\color{blue}{1}&\color{blue}{-1}&\color{blue}{5}&\color{blue}{-9}&\color{orangered}{10} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}-x^{2}+5x-9 } $ with a remainder of $ \color{red}{ 10 } $.