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