Divide using synthetic division $$ ( 2x^{4}+23x^{3}+60x^{2}-125x-500 ) \div ( x+4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-4&2&23&60&-125&-500\\& & -8& -60& 0& \color{black}{500} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{0}&\color{blue}{-125}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+23x^{3}+60x^{2}-125x-500 }{ x+4 } = \color{blue}{2x^{3}+15x^{2}-125} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 4 = 0 $ ( $ x = \color{blue}{ -4 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&23&60&-125&-500\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-4&\color{orangered}{ 2 }&23&60&-125&-500\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 2 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&23&60&-125&-500\\& & \color{blue}{-8} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 23 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}-4&2&\color{orangered}{ 23 }&60&-125&-500\\& & \color{orangered}{-8} & & & \\ \hline &2&\color{orangered}{15}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 15 } = \color{blue}{ -60 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&23&60&-125&-500\\& & -8& \color{blue}{-60} & & \\ \hline &2&\color{blue}{15}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 60 } + \color{orangered}{ \left( -60 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&2&23&\color{orangered}{ 60 }&-125&-500\\& & -8& \color{orangered}{-60} & & \\ \hline &2&15&\color{orangered}{0}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&23&60&-125&-500\\& & -8& -60& \color{blue}{0} & \\ \hline &2&15&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -125 } + \color{orangered}{ 0 } = \color{orangered}{ -125 } $
$$ \begin{array}{c|rrrrr}-4&2&23&60&\color{orangered}{ -125 }&-500\\& & -8& -60& \color{orangered}{0} & \\ \hline &2&15&0&\color{orangered}{-125}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -4 } \cdot \color{blue}{ \left( -125 \right) } = \color{blue}{ 500 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-4}&2&23&60&-125&-500\\& & -8& -60& 0& \color{blue}{500} \\ \hline &2&15&0&\color{blue}{-125}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -500 } + \color{orangered}{ 500 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-4&2&23&60&-125&\color{orangered}{ -500 }\\& & -8& -60& 0& \color{orangered}{500} \\ \hline &\color{blue}{2}&\color{blue}{15}&\color{blue}{0}&\color{blue}{-125}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}+15x^{2}-125 } $ with a remainder of $ \color{red}{ 0 } $.