Divide using synthetic division $$ ( 2x^{4}+10x^{3}-5x^{2}-27x-19 ) \div ( x+5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-5&2&10&-5&-27&-19\\& & -10& 0& 25& \color{black}{10} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{-9} \end{array} $$The solution is:
$$ \frac{ 2x^{4}+10x^{3}-5x^{2}-27x-19 }{ x+5 } = \color{blue}{2x^{3}-5x-2} \color{red}{~-~} \frac{ \color{red}{ 9 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&10&-5&-27&-19\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-5&\color{orangered}{ 2 }&10&-5&-27&-19\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 2 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&10&-5&-27&-19\\& & \color{blue}{-10} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-5&2&\color{orangered}{ 10 }&-5&-27&-19\\& & \color{orangered}{-10} & & & \\ \hline &2&\color{orangered}{0}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&10&-5&-27&-19\\& & -10& \color{blue}{0} & & \\ \hline &2&\color{blue}{0}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 0 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-5&2&10&\color{orangered}{ -5 }&-27&-19\\& & -10& \color{orangered}{0} & & \\ \hline &2&0&\color{orangered}{-5}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 25 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&10&-5&-27&-19\\& & -10& 0& \color{blue}{25} & \\ \hline &2&0&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -27 } + \color{orangered}{ 25 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-5&2&10&-5&\color{orangered}{ -27 }&-19\\& & -10& 0& \color{orangered}{25} & \\ \hline &2&0&-5&\color{orangered}{-2}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -5 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-5}&2&10&-5&-27&-19\\& & -10& 0& 25& \color{blue}{10} \\ \hline &2&0&-5&\color{blue}{-2}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 10 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrrr}-5&2&10&-5&-27&\color{orangered}{ -19 }\\& & -10& 0& 25& \color{orangered}{10} \\ \hline &\color{blue}{2}&\color{blue}{0}&\color{blue}{-5}&\color{blue}{-2}&\color{orangered}{-9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-5x-2 } $ with a remainder of $ \color{red}{ -9 } $.