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