Divide using synthetic division $$ ( 2x^{3}-5x^{2}+6x+8 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-5&6&8\\& & -4& 18& \color{black}{-48} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{24}&\color{orangered}{-40} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-5x^{2}+6x+8 }{ x+2 } = \color{blue}{2x^{2}-9x+24} \color{red}{~-~} \frac{ \color{red}{ 40 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&6&8\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-5&6&8\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&6&8\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -5 }&6&8\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&6&8\\& & -4& \color{blue}{18} & \\ \hline &2&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 18 } = \color{orangered}{ 24 } $
$$ \begin{array}{c|rrrr}-2&2&-5&\color{orangered}{ 6 }&8\\& & -4& \color{orangered}{18} & \\ \hline &2&-9&\color{orangered}{24}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 24 } = \color{blue}{ -48 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-5&6&8\\& & -4& 18& \color{blue}{-48} \\ \hline &2&-9&\color{blue}{24}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -48 \right) } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrr}-2&2&-5&6&\color{orangered}{ 8 }\\& & -4& 18& \color{orangered}{-48} \\ \hline &\color{blue}{2}&\color{blue}{-9}&\color{blue}{24}&\color{orangered}{-40} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-9x+24 } $ with a remainder of $ \color{red}{ -40 } $.