Divide using synthetic division $$ ( -2x^{3}+19x-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-2&0&19&-3\\& & -6& -18& \color{black}{3} \\ \hline &\color{blue}{-2}&\color{blue}{-6}&\color{blue}{1}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ -2x^{3}+19x-3 }{ x-3 } = \color{blue}{-2x^{2}-6x+1} $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&0&19&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -2 }&0&19&-3\\& & & & \\ \hline &\color{orangered}{-2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&0&19&-3\\& & \color{blue}{-6} & & \\ \hline &\color{blue}{-2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrr}3&-2&\color{orangered}{ 0 }&19&-3\\& & \color{orangered}{-6} & & \\ \hline &-2&\color{orangered}{-6}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -6 \right) } = \color{blue}{ -18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&0&19&-3\\& & -6& \color{blue}{-18} & \\ \hline &-2&\color{blue}{-6}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 19 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrr}3&-2&0&\color{orangered}{ 19 }&-3\\& & -6& \color{orangered}{-18} & \\ \hline &-2&-6&\color{orangered}{1}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-2&0&19&-3\\& & -6& -18& \color{blue}{3} \\ \hline &-2&-6&\color{blue}{1}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 3 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}3&-2&0&19&\color{orangered}{ -3 }\\& & -6& -18& \color{orangered}{3} \\ \hline &\color{blue}{-2}&\color{blue}{-6}&\color{blue}{1}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{2}-6x+1 } $ with a remainder of $ \color{red}{ 0 } $.