Divide using synthetic division $$ ( 3x^{3}+5x-1 ) \div ( 3x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&0&5&-1\\& & -6& 12& \color{black}{-34} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{17}&\color{orangered}{-35} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+5x-1 }{ x+2 } = \color{blue}{3x^{2}-6x+17} \color{red}{~-~} \frac{ \color{red}{ 35 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&5&-1\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&0&5&-1\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&5&-1\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ 0 }&5&-1\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-6}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&5&-1\\& & -6& \color{blue}{12} & \\ \hline &3&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 12 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}-2&3&0&\color{orangered}{ 5 }&-1\\& & -6& \color{orangered}{12} & \\ \hline &3&-6&\color{orangered}{17}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 17 } = \color{blue}{ -34 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&0&5&-1\\& & -6& 12& \color{blue}{-34} \\ \hline &3&-6&\color{blue}{17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -34 \right) } = \color{orangered}{ -35 } $
$$ \begin{array}{c|rrrr}-2&3&0&5&\color{orangered}{ -1 }\\& & -6& 12& \color{orangered}{-34} \\ \hline &\color{blue}{3}&\color{blue}{-6}&\color{blue}{17}&\color{orangered}{-35} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-6x+17 } $ with a remainder of $ \color{red}{ -35 } $.