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