Divide using synthetic division $$ ( 2x^{4}-7x^{2}-4x+2 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&2&0&-7&-4&2\\& & -2& 2& 5& \color{black}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{1}&\color{orangered}{1} \end{array} $$The solution is:
$$ \frac{ 2x^{4}-7x^{2}-4x+2 }{ x+1 } = \color{blue}{2x^{3}-2x^{2}-5x+1} ~+~ \frac{ \color{red}{ 1 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&0&-7&-4&2\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ 2 }&0&-7&-4&2\\& & & & & \\ \hline &\color{orangered}{2}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&0&-7&-4&2\\& & \color{blue}{-2} & & & \\ \hline &\color{blue}{2}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -2 \right) } = \color{orangered}{ -2 } $
$$ \begin{array}{c|rrrrr}-1&2&\color{orangered}{ 0 }&-7&-4&2\\& & \color{orangered}{-2} & & & \\ \hline &2&\color{orangered}{-2}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&0&-7&-4&2\\& & -2& \color{blue}{2} & & \\ \hline &2&\color{blue}{-2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 2 } = \color{orangered}{ -5 } $
$$ \begin{array}{c|rrrrr}-1&2&0&\color{orangered}{ -7 }&-4&2\\& & -2& \color{orangered}{2} & & \\ \hline &2&-2&\color{orangered}{-5}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&0&-7&-4&2\\& & -2& 2& \color{blue}{5} & \\ \hline &2&-2&\color{blue}{-5}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 5 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-1&2&0&-7&\color{orangered}{ -4 }&2\\& & -2& 2& \color{orangered}{5} & \\ \hline &2&-2&-5&\color{orangered}{1}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 1 } = \color{blue}{ -1 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&2&0&-7&-4&2\\& & -2& 2& 5& \color{blue}{-1} \\ \hline &2&-2&-5&\color{blue}{1}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ \left( -1 \right) } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}-1&2&0&-7&-4&\color{orangered}{ 2 }\\& & -2& 2& 5& \color{orangered}{-1} \\ \hline &\color{blue}{2}&\color{blue}{-2}&\color{blue}{-5}&\color{blue}{1}&\color{orangered}{1} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{3}-2x^{2}-5x+1 } $ with a remainder of $ \color{red}{ 1 } $.