Divide using synthetic division $$ ( x^{5}-25x^{3}-7x^{2}-37x-18 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&1&0&-25&-7&-37&-18\\& & -5& 25& 0& 35& \color{black}{10} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{-8} \end{array} $$The solution is:
$$ \frac{ x^{5}-25x^{3}-7x^{2}-37x-18 }{ x+5 } = \color{blue}{x^{4}-5x^{3}-7x-2} \color{red}{~-~} \frac{ \color{red}{ 8 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 1 }&0&-25&-7&-37&-18\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & \color{blue}{-5} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}-5&1&\color{orangered}{ 0 }&-25&-7&-37&-18\\& & \color{orangered}{-5} & & & & \\ \hline &1&\color{orangered}{-5}&&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & -5& \color{blue}{25} & & & \\ \hline &1&\color{blue}{-5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -25 } + \color{orangered}{ 25 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&\color{orangered}{ -25 }&-7&-37&-18\\& & -5& \color{orangered}{25} & & & \\ \hline &1&-5&\color{orangered}{0}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & -5& 25& \color{blue}{0} & & \\ \hline &1&-5&\color{blue}{0}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-25&\color{orangered}{ -7 }&-37&-18\\& & -5& 25& \color{orangered}{0} & & \\ \hline &1&-5&0&\color{orangered}{-7}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 35 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & -5& 25& 0& \color{blue}{35} & \\ \hline &1&-5&0&\color{blue}{-7}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -37 } + \color{orangered}{ 35 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-25&-7&\color{orangered}{ -37 }&-18\\& & -5& 25& 0& \color{orangered}{35} & \\ \hline &1&-5&0&-7&\color{orangered}{-2}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-25&-7&-37&-18\\& & -5& 25& 0& 35& \color{blue}{10} \\ \hline &1&-5&0&-7&\color{blue}{-2}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 10 } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-25&-7&-37&\color{orangered}{ -18 }\\& & -5& 25& 0& 35& \color{orangered}{10} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{0}&\color{blue}{-7}&\color{blue}{-2}&\color{orangered}{-8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-5x^{3}-7x-2 } $ with a remainder of $ \color{red}{ -8 } $.