Divide using synthetic division $$ ( 3x^{4}+19x^{3}+20x^{2}-3x-9 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&3&19&20&-3&-9\\& & -15& -20& 0& \color{black}{15} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{6} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+19x^{3}+20x^{2}-3x-9 }{ x+5 } = \color{blue}{3x^{3}+4x^{2}-3} ~+~ \frac{ \color{red}{ 6 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&19&20&-3&-9\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 3 }&19&20&-3&-9\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 3 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&19&20&-3&-9\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-5&3&\color{orangered}{ 19 }&20&-3&-9\\& & \color{orangered}{-15} & & & \\ \hline &3&\color{orangered}{4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 4 } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&19&20&-3&-9\\& & -15& \color{blue}{-20} & & \\ \hline &3&\color{blue}{4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 20 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&3&19&\color{orangered}{ 20 }&-3&-9\\& & -15& \color{orangered}{-20} & & \\ \hline &3&4&\color{orangered}{0}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&19&20&-3&-9\\& & -15& -20& \color{blue}{0} & \\ \hline &3&4&\color{blue}{0}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 0 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-5&3&19&20&\color{orangered}{ -3 }&-9\\& & -15& -20& \color{orangered}{0} & \\ \hline &3&4&0&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&3&19&20&-3&-9\\& & -15& -20& 0& \color{blue}{15} \\ \hline &3&4&0&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 15 } = \color{orangered}{ 6 } $
$$ \begin{array}{c|rrrrr}-5&3&19&20&-3&\color{orangered}{ -9 }\\& & -15& -20& 0& \color{orangered}{15} \\ \hline &\color{blue}{3}&\color{blue}{4}&\color{blue}{0}&\color{blue}{-3}&\color{orangered}{6} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}+4x^{2}-3 } $ with a remainder of $ \color{red}{ 6 } $.