Divide using synthetic division $$ ( -2x^{4}+10x-3 ) \div ( x-3 ) $$
The synthetic division table is:
$$ \begin{array}{c|rrrrr}3&-2&0&0&10&-3\\& & -6& -18& -54& \color{black}{-132} \\ \hline &\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-18}&\color{blue}{-44}&\color{orangered}{-135} \end{array} $$The solution is:
$$ \dfrac{ -2x^{4}+10x-3 }{ x-3 } = \color{blue}{-2x^{3}-6x^{2}-18x-44} \color{red}{~-~} \dfrac{ \color{red}{ 135 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-2&0&0&10&-3\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}3&\color{orangered}{ -2 }&0&0&10&-3\\& & & & & \\ \hline &\color{orangered}{-2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-2&0&0&10&-3\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{-2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -6 } $
$$ \begin{array}{c|rrrrr}3&-2&\color{orangered}{ 0 }&0&10&-3\\& & \color{orangered}{-6} & & & \\ \hline &-2&\color{orangered}{-6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-2&0&0&10&-3\\& & -6& \color{blue}{-18} & & \\ \hline &-2&\color{blue}{-6}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -18 \right) } = \color{orangered}{ -18 } $
$$ \begin{array}{c|rrrrr}3&-2&0&\color{orangered}{ 0 }&10&-3\\& & -6& \color{orangered}{-18} & & \\ \hline &-2&-6&\color{orangered}{-18}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -18 \right) } = \color{blue}{ -54 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-2&0&0&10&-3\\& & -6& -18& \color{blue}{-54} & \\ \hline &-2&-6&\color{blue}{-18}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ \left( -54 \right) } = \color{orangered}{ -44 } $
$$ \begin{array}{c|rrrrr}3&-2&0&0&\color{orangered}{ 10 }&-3\\& & -6& -18& \color{orangered}{-54} & \\ \hline &-2&-6&-18&\color{orangered}{-44}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -44 \right) } = \color{blue}{ -132 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{3}&-2&0&0&10&-3\\& & -6& -18& -54& \color{blue}{-132} \\ \hline &-2&-6&-18&\color{blue}{-44}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -132 \right) } = \color{orangered}{ -135 } $
$$ \begin{array}{c|rrrrr}3&-2&0&0&10&\color{orangered}{ -3 }\\& & -6& -18& -54& \color{orangered}{-132} \\ \hline &\color{blue}{-2}&\color{blue}{-6}&\color{blue}{-18}&\color{blue}{-44}&\color{orangered}{-135} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -2x^{3}-6x^{2}-18x-44 } $ with a remainder of $ \color{red}{ -135 } $.