Divide using synthetic division $$ ( 12x^{3}-x^{2}-7x+2 ) \div ( x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-2&12&-1&-7&2\\& & -24& 50& \color{black}{-86} \\ \hline &\color{blue}{12}&\color{blue}{-25}&\color{blue}{43}&\color{orangered}{-84} \end{array} $$The solution is:
$$ \frac{ 12x^{3}-x^{2}-7x+2 }{ x+2 } = \color{blue}{12x^{2}-25x+43} \color{red}{~-~} \frac{ \color{red}{ 84 } }{ x+2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 2 = 0 $ ( $ x = \color{blue}{ -2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&-1&-7&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-2&\color{orangered}{ 12 }&-1&-7&2\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 12 } = \color{blue}{ -24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&-1&-7&2\\& & \color{blue}{-24} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ \left( -24 \right) } = \color{orangered}{ -25 } $
$$ \begin{array}{c|rrrr}-2&12&\color{orangered}{ -1 }&-7&2\\& & \color{orangered}{-24} & & \\ \hline &12&\color{orangered}{-25}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -25 \right) } = \color{blue}{ 50 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&-1&-7&2\\& & -24& \color{blue}{50} & \\ \hline &12&\color{blue}{-25}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 50 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrrr}-2&12&-1&\color{orangered}{ -7 }&2\\& & -24& \color{orangered}{50} & \\ \hline &12&-25&\color{orangered}{43}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 43 } = \color{blue}{ -86 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-2}&12&-1&-7&2\\& & -24& 50& \color{blue}{-86} \\ \hline &12&-25&\color{blue}{43}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -86 \right) } = \color{orangered}{ -84 } $
$$ \begin{array}{c|rrrr}-2&12&-1&-7&\color{orangered}{ 2 }\\& & -24& 50& \color{orangered}{-86} \\ \hline &\color{blue}{12}&\color{blue}{-25}&\color{blue}{43}&\color{orangered}{-84} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}-25x+43 } $ with a remainder of $ \color{red}{ -84 } $.