Divide using synthetic division $$ ( 3x^{4}+4x^{3}-6x^{2}-7x+2 ) \div ( -12x+2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-2&3&4&-6&-7&2\\& & -6& 4& 4& \color{black}{6} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-3}&\color{orangered}{8} \end{array} $$The solution is:
$$ \frac{ 3x^{4}+4x^{3}-6x^{2}-7x+2 }{ x+2 } = \color{blue}{3x^{3}-2x^{2}-2x-3} ~+~ \frac{ \color{red}{ 8 } }{ x+2 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&4&-6&-7&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-2&\color{orangered}{ 3 }&4&-6&-7&2\\& & & & & \\ \hline &\color{orangered}{3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ 3 } = \color{blue}{ -6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&4&-6&-7&2\\& & \color{blue}{-6} & & & \\ \hline &\color{blue}{3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ \left( -6 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-2&3&\color{orangered}{ 4 }&-6&-7&2\\& & \color{orangered}{-6} & & & \\ \hline &3&\color{orangered}{-2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&4&-6&-7&2\\& & -6& \color{blue}{4} & & \\ \hline &3&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 4 } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-2&3&4&\color{orangered}{ -6 }&-7&2\\& & -6& \color{orangered}{4} & & \\ \hline &3&-2&\color{orangered}{-2}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -2 \right) } = \color{blue}{ 4 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&4&-6&-7&2\\& & -6& 4& \color{blue}{4} & \\ \hline &3&-2&\color{blue}{-2}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 4 } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-2&3&4&-6&\color{orangered}{ -7 }&2\\& & -6& 4& \color{orangered}{4} & \\ \hline &3&-2&-2&\color{orangered}{-3}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -2 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-2}&3&4&-6&-7&2\\& & -6& 4& 4& \color{blue}{6} \\ \hline &3&-2&-2&\color{blue}{-3}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 6 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}-2&3&4&-6&-7&\color{orangered}{ 2 }\\& & -6& 4& 4& \color{orangered}{6} \\ \hline &\color{blue}{3}&\color{blue}{-2}&\color{blue}{-2}&\color{blue}{-3}&\color{orangered}{8} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{3}-2x^{2}-2x-3 } $ with a remainder of $ \color{red}{ 8 } $.