Divide using synthetic division $$ ( 3x^{3}-2x^{2}+x-5 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&3&-2&1&-5\\& & -6& 16& \color{black}{-34} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{17}&\color{orangered}{-39} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-2x^{2}+x-5 }{ x+2 } = \color{blue}{3x^{2}-8x+17} \color{red}{~-~} \frac{ \color{red}{ 39 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-2&1&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 3 }&-2&1&-5\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-2&1&-5\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-2&3&\color{orangered}{ -2 }&1&-5\\& & \color{orangered}{-6} & & \\ \hline &3&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 16 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-2&1&-5\\& & -6& \color{blue}{16} & \\ \hline &3&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 16 } = \color{orangered}{ 17 } $
$$ \begin{array}{c|rrrr}-2&3&-2&\color{orangered}{ 1 }&-5\\& & -6& \color{orangered}{16} & \\ \hline &3&-8&\color{orangered}{17}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&3&-2&1&-5\\& & -6& 16& \color{blue}{-34} \\ \hline &3&-8&\color{blue}{17}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -34 \right) } = \color{orangered}{ -39 } $
$$ \begin{array}{c|rrrr}-2&3&-2&1&\color{orangered}{ -5 }\\& & -6& 16& \color{orangered}{-34} \\ \hline &\color{blue}{3}&\color{blue}{-8}&\color{blue}{17}&\color{orangered}{-39} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}-8x+17 } $ with a remainder of $ \color{red}{ -39 } $.