Divide using synthetic division $$ ( 2x^{3}+14x^{2}+19x-8 ) \div ( x+5 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrr}-5&2&14&19&-8\\& & -10& -20& \color{black}{5} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$The solution is:
$$ \dfrac{ 2x^{3}+14x^{2}+19x-8 }{ x+5 } = \color{blue}{2x^{2}+4x-1} \color{red}{~-~} \dfrac{ \color{red}{ 3 } }{ x+5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&14&19&-8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-5&\color{orangered}{ 2 }&14&19&-8\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&14&19&-8\\& & \color{blue}{-10} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}-5&2&\color{orangered}{ 14 }&19&-8\\& & \color{orangered}{-10} & & \\ \hline &2&\color{orangered}{4}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&14&19&-8\\& & -10& \color{blue}{-20} & \\ \hline &2&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -20 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}-5&2&14&\color{orangered}{ 19 }&-8\\& & -10& \color{orangered}{-20} & \\ \hline &2&4&\color{orangered}{-1}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-5}&2&14&19&-8\\& & -10& -20& \color{blue}{5} \\ \hline &2&4&\color{blue}{-1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -8 } + \color{orangered}{ 5 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrr}-5&2&14&19&\color{orangered}{ -8 }\\& & -10& -20& \color{orangered}{5} \\ \hline &\color{blue}{2}&\color{blue}{4}&\color{blue}{-1}&\color{orangered}{-3} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+4x-1 } $ with a remainder of $ \color{red}{ -3 } $.