Divide using synthetic division $$ ( x^{5}-24x^{3}-19x+37 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrrr}-5&1&0&-24&0&-19&37\\& & -5& 25& -5& 25& \color{black}{-30} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{7} \end{array} $$The solution is:
$$ \frac{ x^{5}-24x^{3}-19x+37 }{ x+5 } = \color{blue}{x^{4}-5x^{3}+x^{2}-5x+6} ~+~ \frac{ \color{red}{ 7 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & & & & & \\ \hline &&&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrrr}-5&\color{orangered}{ 1 }&0&-24&0&-19&37\\& & & & & & \\ \hline &\color{orangered}{1}&&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & \color{blue}{-5} & & & & \\ \hline &\color{blue}{1}&&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}-5&1&\color{orangered}{ 0 }&-24&0&-19&37\\& & \color{orangered}{-5} & & & & \\ \hline &1&\color{orangered}{-5}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & -5& \color{blue}{25} & & & \\ \hline &1&\color{blue}{-5}&&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 25 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&\color{orangered}{ -24 }&0&-19&37\\& & -5& \color{orangered}{25} & & & \\ \hline &1&-5&\color{orangered}{1}&&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 1 } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & -5& 25& \color{blue}{-5} & & \\ \hline &1&-5&\color{blue}{1}&&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-24&\color{orangered}{ 0 }&-19&37\\& & -5& 25& \color{orangered}{-5} & & \\ \hline &1&-5&1&\color{orangered}{-5}&& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & -5& 25& -5& \color{blue}{25} & \\ \hline &1&-5&1&\color{blue}{-5}&& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 25 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-24&0&\color{orangered}{ -19 }&37\\& & -5& 25& -5& \color{orangered}{25} & \\ \hline &1&-5&1&-5&\color{orangered}{6}& \end{array} $$Step 10 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 6 } = \color{blue}{ -30 } $.
$$ \begin{array}{c|rrrrrr}\color{blue}{-5}&1&0&-24&0&-19&37\\& & -5& 25& -5& 25& \color{blue}{-30} \\ \hline &1&-5&1&-5&\color{blue}{6}& \end{array} $$Step 11 : Add down last column: $ \color{orangered}{ 37 } + \color{orangered}{ \left( -30 \right) } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrrrr}-5&1&0&-24&0&-19&\color{orangered}{ 37 }\\& & -5& 25& -5& 25& \color{orangered}{-30} \\ \hline &\color{blue}{1}&\color{blue}{-5}&\color{blue}{1}&\color{blue}{-5}&\color{blue}{6}&\color{orangered}{7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{4}-5x^{3}+x^{2}-5x+6 } $ with a remainder of $ \color{red}{ 7 } $.