Divide using synthetic division $$ ( 2x^{3}+5x^{2}-12x-2 ) \div ( x-1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}1&2&5&-12&-2\\& & 2& 7& \color{black}{-5} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-5}&\color{orangered}{-7} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+5x^{2}-12x-2 }{ x-1 } = \color{blue}{2x^{2}+7x-5} \color{red}{~-~} \frac{ \color{red}{ 7 } }{ x-1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&5&-12&-2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}1&\color{orangered}{ 2 }&5&-12&-2\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 2 } = \color{blue}{ 2 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&5&-12&-2\\& & \color{blue}{2} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 2 } = \color{orangered}{ 7 } $
$$ \begin{array}{c|rrrr}1&2&\color{orangered}{ 5 }&-12&-2\\& & \color{orangered}{2} & & \\ \hline &2&\color{orangered}{7}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ 7 } = \color{blue}{ 7 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&5&-12&-2\\& & 2& \color{blue}{7} & \\ \hline &2&\color{blue}{7}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 7 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrr}1&2&5&\color{orangered}{ -12 }&-2\\& & 2& \color{orangered}{7} & \\ \hline &2&7&\color{orangered}{-5}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 1 } \cdot \color{blue}{ \left( -5 \right) } = \color{blue}{ -5 } $.
$$ \begin{array}{c|rrrr}\color{blue}{1}&2&5&-12&-2\\& & 2& 7& \color{blue}{-5} \\ \hline &2&7&\color{blue}{-5}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -2 } + \color{orangered}{ \left( -5 \right) } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}1&2&5&-12&\color{orangered}{ -2 }\\& & 2& 7& \color{orangered}{-5} \\ \hline &\color{blue}{2}&\color{blue}{7}&\color{blue}{-5}&\color{orangered}{-7} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+7x-5 } $ with a remainder of $ \color{red}{ -7 } $.