Divide using synthetic division $$ ( -2x^{3}+5x^{2}-x+3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-2&5&-1&3\\& & -6& -3& \color{black}{-12} \\ \hline &\color{blue}{-2}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{-9} \end{array} $$The solution is:
$$ \frac{ -2x^{3}+5x^{2}-x+3 }{ x-3 } = \color{blue}{-2x^{2}-x-4} \color{red}{~-~} \frac{ \color{red}{ 9 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&5&-1&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -2 }&5&-1&3\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&5&-1&3\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}3&-2&\color{orangered}{ 5 }&-1&3\\& & \color{orangered}{-6} & & \\ \hline &-2&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&5&-1&3\\& & -6& \color{blue}{-3} & \\ \hline &-2&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}3&-2&5&\color{orangered}{ -1 }&3\\& & -6& \color{orangered}{-3} & \\ \hline &-2&-1&\color{orangered}{-4}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ -12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&5&-1&3\\& & -6& -3& \color{blue}{-12} \\ \hline &-2&-1&\color{blue}{-4}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ \left( -12 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}3&-2&5&-1&\color{orangered}{ 3 }\\& & -6& -3& \color{orangered}{-12} \\ \hline &\color{blue}{-2}&\color{blue}{-1}&\color{blue}{-4}&\color{orangered}{-9} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{2}-x-4 } $ with a remainder of $ \color{red}{ -9 } $.