Divide using synthetic division $$ ( 2x^{4}-2x^{3}-19x^{2}-24x-24 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&-2&-19&-24&-24\\& & -4& 12& 14& \color{black}{20} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{-7}&\color{blue}{-10}&\color{orangered}{-4} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-2x^{3}-19x^{2}-24x-24 }{ x+2 } = \color{blue}{2x^{3}-6x^{2}-7x-10} \color{red}{~-~} \frac{ \color{red}{ 4 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-19&-24&-24\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&-2&-19&-24&-24\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-19&-24&-24\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ -2 }&-19&-24&-24\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{-6}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-19&-24&-24\\& & -4& \color{blue}{12} & & \\ \hline &2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 12 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&\color{orangered}{ -19 }&-24&-24\\& & -4& \color{orangered}{12} & & \\ \hline &2&-6&\color{orangered}{-7}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -7 \right) } = \color{blue}{ 14 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-19&-24&-24\\& & -4& 12& \color{blue}{14} & \\ \hline &2&-6&\color{blue}{-7}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 14 } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&-19&\color{orangered}{ -24 }&-24\\& & -4& 12& \color{orangered}{14} & \\ \hline &2&-6&-7&\color{orangered}{-10}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 20 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-19&-24&-24\\& & -4& 12& 14& \color{blue}{20} \\ \hline &2&-6&-7&\color{blue}{-10}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -24 } + \color{orangered}{ 20 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&-19&-24&\color{orangered}{ -24 }\\& & -4& 12& 14& \color{orangered}{20} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{-7}&\color{blue}{-10}&\color{orangered}{-4} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-6x^{2}-7x-10 } $ with a remainder of $ \color{red}{ -4 } $.