Divide using synthetic division $$ ( 2x^{3}-x^{2}+8x+5 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-1&8&5\\& & -4& 10& \color{black}{-36} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{18}&\color{orangered}{-31} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-x^{2}+8x+5 }{ x+2 } = \color{blue}{2x^{2}-5x+18} \color{red}{~-~} \frac{ \color{red}{ 31 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&8&5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-1&8&5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 2 } = \color{blue}{ -4 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&8&5\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -1 }&8&5\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&8&5\\& & -4& \color{blue}{10} & \\ \hline &2&\color{blue}{-5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 10 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}-2&2&-1&\color{orangered}{ 8 }&5\\& & -4& \color{orangered}{10} & \\ \hline &2&-5&\color{orangered}{18}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 18 } = \color{blue}{ -36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-1&8&5\\& & -4& 10& \color{blue}{-36} \\ \hline &2&-5&\color{blue}{18}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -36 \right) } = \color{orangered}{ -31 } $
$$ \begin{array}{c|rrrr}-2&2&-1&8&\color{orangered}{ 5 }\\& & -4& 10& \color{orangered}{-36} \\ \hline &\color{blue}{2}&\color{blue}{-5}&\color{blue}{18}&\color{orangered}{-31} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-5x+18 } $ with a remainder of $ \color{red}{ -31 } $.