Divide using synthetic division $$ ( 2x^{3}-2x^{2}+4x-5 ) \div ( 2x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&-2&4&-5\\& & -6& 24& \color{black}{-84} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{28}&\color{orangered}{-89} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-2x^{2}+4x-5 }{ x+3 } = \color{blue}{2x^{2}-8x+28} \color{red}{~-~} \frac{ \color{red}{ 89 } }{ x+3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-2&4&-5\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&-2&4&-5\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 2 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-2&4&-5\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -8 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ -2 }&4&-5\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-8}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -8 \right) } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-2&4&-5\\& & -6& \color{blue}{24} & \\ \hline &2&\color{blue}{-8}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 24 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}-3&2&-2&\color{orangered}{ 4 }&-5\\& & -6& \color{orangered}{24} & \\ \hline &2&-8&\color{orangered}{28}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 28 } = \color{blue}{ -84 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-2&4&-5\\& & -6& 24& \color{blue}{-84} \\ \hline &2&-8&\color{blue}{28}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ \left( -84 \right) } = \color{orangered}{ -89 } $
$$ \begin{array}{c|rrrr}-3&2&-2&4&\color{orangered}{ -5 }\\& & -6& 24& \color{orangered}{-84} \\ \hline &\color{blue}{2}&\color{blue}{-8}&\color{blue}{28}&\color{orangered}{-89} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-8x+28 } $ with a remainder of $ \color{red}{ -89 } $.