Divide using synthetic division $$ ( 3x^{3}-6x^{2}+8x+17 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&-6&8&17\\& & -6& 24& \color{black}{-64} \\ \hline &\color{blue}{3}&\color{blue}{-12}&\color{blue}{32}&\color{orangered}{-47} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-6x^{2}+8x+17 }{ x+2 } = \color{blue}{3x^{2}-12x+32} \color{red}{~-~} \frac{ \color{red}{ 47 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-6&8&17\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&-6&8&17\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-6&8&17\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ -6 }&8&17\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-12}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-6&8&17\\& & -6& \color{blue}{24} & \\ \hline &3&\color{blue}{-12}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 24 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}-2&3&-6&\color{orangered}{ 8 }&17\\& & -6& \color{orangered}{24} & \\ \hline &3&-12&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 32 } = \color{blue}{ -64 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-6&8&17\\& & -6& 24& \color{blue}{-64} \\ \hline &3&-12&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 17 } + \color{orangered}{ \left( -64 \right) } = \color{orangered}{ -47 } $
$$ \begin{array}{c|rrrr}-2&3&-6&8&\color{orangered}{ 17 }\\& & -6& 24& \color{orangered}{-64} \\ \hline &\color{blue}{3}&\color{blue}{-12}&\color{blue}{32}&\color{orangered}{-47} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-12x+32 } $ with a remainder of $ \color{red}{ -47 } $.