Divide using synthetic division $$ ( 2x^{3}-4x^{2}-5x+3 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-3&2&-4&-5&3\\& & -6& 30& \color{black}{-75} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{25}&\color{orangered}{-72} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-4x^{2}-5x+3 }{ x+3 } = \color{blue}{2x^{2}-10x+25} \color{red}{~-~} \frac{ \color{red}{ 72 } }{ x+3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&-5&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-3&\color{orangered}{ 2 }&-4&-5&3\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&-5&3\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrr}-3&2&\color{orangered}{ -4 }&-5&3\\& & \color{orangered}{-6} & & \\ \hline &2&\color{orangered}{-10}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&-5&3\\& & -6& \color{blue}{30} & \\ \hline &2&\color{blue}{-10}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 30 } = \color{orangered}{ 25 } $
$$ \begin{array}{c|rrrr}-3&2&-4&\color{orangered}{ -5 }&3\\& & -6& \color{orangered}{30} & \\ \hline &2&-10&\color{orangered}{25}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 25 } = \color{blue}{ -75 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-3}&2&-4&-5&3\\& & -6& 30& \color{blue}{-75} \\ \hline &2&-10&\color{blue}{25}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -75 \right) } = \color{orangered}{ -72 } $
$$ \begin{array}{c|rrrr}-3&2&-4&-5&\color{orangered}{ 3 }\\& & -6& 30& \color{orangered}{-75} \\ \hline &\color{blue}{2}&\color{blue}{-10}&\color{blue}{25}&\color{orangered}{-72} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-10x+25 } $ with a remainder of $ \color{red}{ -72 } $.