Divide using synthetic division $$ ( -4x^{4}+19x^{3}+3x^{2}+7x+15 ) \div ( x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}5&-4&19&3&7&15\\& & -20& -5& -10& \color{black}{-15} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -4x^{4}+19x^{3}+3x^{2}+7x+15 }{ x-5 } = \color{blue}{-4x^{3}-x^{2}-2x-3} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-4&19&3&7&15\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}5&\color{orangered}{ -4 }&19&3&7&15\\& & & & & \\ \hline &\color{orangered}{-4}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-4&19&3&7&15\\& & \color{blue}{-20} & & & \\ \hline &\color{blue}{-4}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}5&-4&\color{orangered}{ 19 }&3&7&15\\& & \color{orangered}{-20} & & & \\ \hline &-4&\color{orangered}{-1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-4&19&3&7&15\\& & -20& \color{blue}{-5} & & \\ \hline &-4&\color{blue}{-1}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}5&-4&19&\color{orangered}{ 3 }&7&15\\& & -20& \color{orangered}{-5} & & \\ \hline &-4&-1&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-4&19&3&7&15\\& & -20& -5& \color{blue}{-10} & \\ \hline &-4&-1&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}5&-4&19&3&\color{orangered}{ 7 }&15\\& & -20& -5& \color{orangered}{-10} & \\ \hline &-4&-1&-2&\color{orangered}{-3}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{5}&-4&19&3&7&15\\& & -20& -5& -10& \color{blue}{-15} \\ \hline &-4&-1&-2&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 15 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}5&-4&19&3&7&\color{orangered}{ 15 }\\& & -20& -5& -10& \color{orangered}{-15} \\ \hline &\color{blue}{-4}&\color{blue}{-1}&\color{blue}{-2}&\color{blue}{-3}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -4x^{3}-x^{2}-2x-3 } $ with a remainder of $ \color{red}{ 0 } $.