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