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