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