Divide using synthetic division $$ ( 9x^{3}+5x-3 ) \div ( 3x+1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}-1&9&0&5&-3\\& & -9& 9& \color{black}{-14} \\ \hline &\color{blue}{9}&\color{blue}{-9}&\color{blue}{14}&\color{orangered}{-17} \end{array} $$The solution is:
$$ \frac{ 9x^{3}+5x-3 }{ x+1 } = \color{blue}{9x^{2}-9x+14} \color{red}{~-~} \frac{ \color{red}{ 17 } }{ x+1 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&9&0&5&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}-1&\color{orangered}{ 9 }&0&5&-3\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 9 } = \color{blue}{ -9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&9&0&5&-3\\& & \color{blue}{-9} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ \left( -9 \right) } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}-1&9&\color{orangered}{ 0 }&5&-3\\& & \color{orangered}{-9} & & \\ \hline &9&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&9&0&5&-3\\& & -9& \color{blue}{9} & \\ \hline &9&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 9 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}-1&9&0&\color{orangered}{ 5 }&-3\\& & -9& \color{orangered}{9} & \\ \hline &9&-9&\color{orangered}{14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -1 } \cdot \color{blue}{ 14 } = \color{blue}{ -14 } $.
$$ \begin{array}{c|rrrr}\color{blue}{-1}&9&0&5&-3\\& & -9& 9& \color{blue}{-14} \\ \hline &9&-9&\color{blue}{14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ \left( -14 \right) } = \color{orangered}{ -17 } $
$$ \begin{array}{c|rrrr}-1&9&0&5&\color{orangered}{ -3 }\\& & -9& 9& \color{orangered}{-14} \\ \hline &\color{blue}{9}&\color{blue}{-9}&\color{blue}{14}&\color{orangered}{-17} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}-9x+14 } $ with a remainder of $ \color{red}{ -17 } $.