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