Divide using synthetic division $$ ( 2x^{4}-2x^{3}-13x^{2}+2x+8 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&2&-2&-13&2&8\\& & -4& 12& 2& \color{black}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-2x^{3}-13x^{2}+2x+8 }{ x+2 } = \color{blue}{2x^{3}-6x^{2}-x+4} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-13&2&8\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 2 }&-2&-13&2&8\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-13&2&8\\& & \color{blue}{-4} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}-2&2&\color{orangered}{ -2 }&-13&2&8\\& & \color{orangered}{-4} & & & \\ \hline &2&\color{orangered}{-6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-13&2&8\\& & -4& \color{blue}{12} & & \\ \hline &2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -13 } + \color{orangered}{ 12 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&\color{orangered}{ -13 }&2&8\\& & -4& \color{orangered}{12} & & \\ \hline &2&-6&\color{orangered}{-1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-13&2&8\\& & -4& 12& \color{blue}{2} & \\ \hline &2&-6&\color{blue}{-1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 2 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&-13&\color{orangered}{ 2 }&8\\& & -4& 12& \color{orangered}{2} & \\ \hline &2&-6&-1&\color{orangered}{4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&2&-2&-13&2&8\\& & -4& 12& 2& \color{blue}{-8} \\ \hline &2&-6&-1&\color{blue}{4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&2&-2&-13&2&\color{orangered}{ 8 }\\& & -4& 12& 2& \color{orangered}{-8} \\ \hline &\color{blue}{2}&\color{blue}{-6}&\color{blue}{-1}&\color{blue}{4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-6x^{2}-x+4 } $ with a remainder of $ \color{red}{ 0 } $.