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