Divide using synthetic division $$ ( 2x^{3}-4x^{2}+5x-2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&2&-4&5&-2\\& & -4& 16& \color{black}{-42} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{21}&\color{orangered}{-44} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-4x^{2}+5x-2 }{ x+2 } = \color{blue}{2x^{2}-8x+21} \color{red}{~-~} \frac{ \color{red}{ 44 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-4&5&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 2 }&-4&5&-2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-4&5&-2\\& & \color{blue}{-4} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -4 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-2&2&\color{orangered}{ -4 }&5&-2\\& & \color{orangered}{-4} & & \\ \hline &2&\color{orangered}{-8}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-4&5&-2\\& & -4& \color{blue}{16} & \\ \hline &2&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 16 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}-2&2&-4&\color{orangered}{ 5 }&-2\\& & -4& \color{orangered}{16} & \\ \hline &2&-8&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 21 } = \color{blue}{ -42 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&2&-4&5&-2\\& & -4& 16& \color{blue}{-42} \\ \hline &2&-8&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -42 \right) } = \color{orangered}{ -44 } $
$$ \begin{array}{c|rrrr}-2&2&-4&5&\color{orangered}{ -2 }\\& & -4& 16& \color{orangered}{-42} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{21}&\color{orangered}{-44} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-8x+21 } $ with a remainder of $ \color{red}{ -44 } $.