Divide using synthetic division $$ ( 3x^{2}+7x-18 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&3&7&-18\\& & 9& \color{black}{48} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{orangered}{30} \end{array} $$The solution is:
$$ \frac{ 3x^{2}+7x-18 }{ x-3 } = \color{blue}{3x+16} ~+~ \frac{ \color{red}{ 30 } }{ x-3 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -3 = 0 $ ( $ x = \color{blue}{ 3 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&7&-18\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 3 }&7&-18\\& & & \\ \hline &\color{orangered}{3}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&7&-18\\& & \color{blue}{9} & \\ \hline &\color{blue}{3}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 9 } = \color{orangered}{ 16 } $
$$ \begin{array}{c|rrr}3&3&\color{orangered}{ 7 }&-18\\& & \color{orangered}{9} & \\ \hline &3&\color{orangered}{16}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 16 } = \color{blue}{ 48 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&3&7&-18\\& & 9& \color{blue}{48} \\ \hline &3&\color{blue}{16}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -18 } + \color{orangered}{ 48 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrr}3&3&7&\color{orangered}{ -18 }\\& & 9& \color{orangered}{48} \\ \hline &\color{blue}{3}&\color{blue}{16}&\color{orangered}{30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x+16 } $ with a remainder of $ \color{red}{ 30 } $.