Divide using synthetic division $$ ( 6x^{4}+8x^{3}-4x^{2}+17x+18 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&6&8&-4&17&18\\& & -12& 8& -8& \color{black}{-18} \\ \hline &\color{blue}{6}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{9}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 6x^{4}+8x^{3}-4x^{2}+17x+18 }{ x+2 } = \color{blue}{6x^{3}-4x^{2}+4x+9} $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&8&-4&17&18\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 6 }&8&-4&17&18\\& & & & & \\ \hline &\color{orangered}{6}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&8&-4&17&18\\& & \color{blue}{-12} & & & \\ \hline &\color{blue}{6}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-2&6&\color{orangered}{ 8 }&-4&17&18\\& & \color{orangered}{-12} & & & \\ \hline &6&\color{orangered}{-4}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&8&-4&17&18\\& & -12& \color{blue}{8} & & \\ \hline &6&\color{blue}{-4}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 8 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrrr}-2&6&8&\color{orangered}{ -4 }&17&18\\& & -12& \color{orangered}{8} & & \\ \hline &6&-4&\color{orangered}{4}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 4 } = \color{blue}{ -8 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&8&-4&17&18\\& & -12& 8& \color{blue}{-8} & \\ \hline &6&-4&\color{blue}{4}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -8 \right) } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrrr}-2&6&8&-4&\color{orangered}{ 17 }&18\\& & -12& 8& \color{orangered}{-8} & \\ \hline &6&-4&4&\color{orangered}{9}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 9 } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&6&8&-4&17&18\\& & -12& 8& -8& \color{blue}{-18} \\ \hline &6&-4&4&\color{blue}{9}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 18 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-2&6&8&-4&17&\color{orangered}{ 18 }\\& & -12& 8& -8& \color{orangered}{-18} \\ \hline &\color{blue}{6}&\color{blue}{-4}&\color{blue}{4}&\color{blue}{9}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{3}-4x^{2}+4x+9 } $ with a remainder of $ \color{red}{ 0 } $.