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