Divide using synthetic division $$ ( -3x^{4}+7x^{3}+5x-7 ) \div ( x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-1&-3&7&0&5&-7\\& & 3& -10& 10& \color{black}{-15} \\ \hline &\color{blue}{-3}&\color{blue}{10}&\color{blue}{-10}&\color{blue}{15}&\color{orangered}{-22} \end{array} $$The solution is:
$$ \frac{ -3x^{4}+7x^{3}+5x-7 }{ x+1 } = \color{blue}{-3x^{3}+10x^{2}-10x+15} \color{red}{~-~} \frac{ \color{red}{ 22 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&-3&7&0&5&-7\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-1&\color{orangered}{ -3 }&7&0&5&-7\\& & & & & \\ \hline &\color{orangered}{-3}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&-3&7&0&5&-7\\& & \color{blue}{3} & & & \\ \hline &\color{blue}{-3}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 3 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrrr}-1&-3&\color{orangered}{ 7 }&0&5&-7\\& & \color{orangered}{3} & & & \\ \hline &-3&\color{orangered}{10}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 10 } = \color{blue}{ -10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&-3&7&0&5&-7\\& & 3& \color{blue}{-10} & & \\ \hline &-3&\color{blue}{10}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -10 \right) } = \color{orangered}{ -10 } $
$$ \begin{array}{c|rrrrr}-1&-3&7&\color{orangered}{ 0 }&5&-7\\& & 3& \color{orangered}{-10} & & \\ \hline &-3&10&\color{orangered}{-10}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -10 \right) } = \color{blue}{ 10 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&-3&7&0&5&-7\\& & 3& -10& \color{blue}{10} & \\ \hline &-3&10&\color{blue}{-10}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 10 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrrr}-1&-3&7&0&\color{orangered}{ 5 }&-7\\& & 3& -10& \color{orangered}{10} & \\ \hline &-3&10&-10&\color{orangered}{15}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 15 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-1}&-3&7&0&5&-7\\& & 3& -10& 10& \color{blue}{-15} \\ \hline &-3&10&-10&\color{blue}{15}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -22 } $
$$ \begin{array}{c|rrrrr}-1&-3&7&0&5&\color{orangered}{ -7 }\\& & 3& -10& 10& \color{orangered}{-15} \\ \hline &\color{blue}{-3}&\color{blue}{10}&\color{blue}{-10}&\color{blue}{15}&\color{orangered}{-22} \end{array} $$Bottom line represents the quotient $ \color{blue}{ -3x^{3}+10x^{2}-10x+15 } $ with a remainder of $ \color{red}{ -22 } $.