Divide using synthetic division $$ ( 9x^{3}-7x-15 ) \div ( 3x-2 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}2&9&0&-7&-15\\& & 18& 36& \color{black}{58} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{29}&\color{orangered}{43} \end{array} $$The solution is:
$$ \frac{ 9x^{3}-7x-15 }{ x-2 } = \color{blue}{9x^{2}+18x+29} ~+~ \frac{ \color{red}{ 43 } }{ x-2 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -2 = 0 $ ( $ x = \color{blue}{ 2 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&0&-7&-15\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}2&\color{orangered}{ 9 }&0&-7&-15\\& & & & \\ \hline &\color{orangered}{9}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 9 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&0&-7&-15\\& & \color{blue}{18} & & \\ \hline &\color{blue}{9}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 18 } = \color{orangered}{ 18 } $
$$ \begin{array}{c|rrrr}2&9&\color{orangered}{ 0 }&-7&-15\\& & \color{orangered}{18} & & \\ \hline &9&\color{orangered}{18}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 18 } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&0&-7&-15\\& & 18& \color{blue}{36} & \\ \hline &9&\color{blue}{18}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 36 } = \color{orangered}{ 29 } $
$$ \begin{array}{c|rrrr}2&9&0&\color{orangered}{ -7 }&-15\\& & 18& \color{orangered}{36} & \\ \hline &9&18&\color{orangered}{29}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 2 } \cdot \color{blue}{ 29 } = \color{blue}{ 58 } $.
$$ \begin{array}{c|rrrr}\color{blue}{2}&9&0&-7&-15\\& & 18& 36& \color{blue}{58} \\ \hline &9&18&\color{blue}{29}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -15 } + \color{orangered}{ 58 } = \color{orangered}{ 43 } $
$$ \begin{array}{c|rrrr}2&9&0&-7&\color{orangered}{ -15 }\\& & 18& 36& \color{orangered}{58} \\ \hline &\color{blue}{9}&\color{blue}{18}&\color{blue}{29}&\color{orangered}{43} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 9x^{2}+18x+29 } $ with a remainder of $ \color{red}{ 43 } $.