Divide using synthetic division $$ ( -x^{3}-39x^{2}+31x+70 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&-1&-39&31&70\\& & -3& -126& \color{black}{-285} \\ \hline &\color{blue}{-1}&\color{blue}{-42}&\color{blue}{-95}&\color{orangered}{-215} \end{array} $$The solution is:
$$ \frac{ -x^{3}-39x^{2}+31x+70 }{ x-3 } = \color{blue}{-x^{2}-42x-95} \color{red}{~-~} \frac{ \color{red}{ 215 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-1&-39&31&70\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ -1 }&-39&31&70\\& & & & \\ \hline &\color{orangered}{-1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ -3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-1&-39&31&70\\& & \color{blue}{-3} & & \\ \hline &\color{blue}{-1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -39 } + \color{orangered}{ \left( -3 \right) } = \color{orangered}{ -42 } $
$$ \begin{array}{c|rrrr}3&-1&\color{orangered}{ -39 }&31&70\\& & \color{orangered}{-3} & & \\ \hline &-1&\color{orangered}{-42}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -42 \right) } = \color{blue}{ -126 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-1&-39&31&70\\& & -3& \color{blue}{-126} & \\ \hline &-1&\color{blue}{-42}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 31 } + \color{orangered}{ \left( -126 \right) } = \color{orangered}{ -95 } $
$$ \begin{array}{c|rrrr}3&-1&-39&\color{orangered}{ 31 }&70\\& & -3& \color{orangered}{-126} & \\ \hline &-1&-42&\color{orangered}{-95}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ \left( -95 \right) } = \color{blue}{ -285 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&-1&-39&31&70\\& & -3& -126& \color{blue}{-285} \\ \hline &-1&-42&\color{blue}{-95}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 70 } + \color{orangered}{ \left( -285 \right) } = \color{orangered}{ -215 } $
$$ \begin{array}{c|rrrr}3&-1&-39&31&\color{orangered}{ 70 }\\& & -3& -126& \color{orangered}{-285} \\ \hline &\color{blue}{-1}&\color{blue}{-42}&\color{blue}{-95}&\color{orangered}{-215} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -x^{2}-42x-95 } $ with a remainder of $ \color{red}{ -215 } $.