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