Divide using synthetic division $$ ( 3x^{3}-4x^{2}+7 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&-4&0&7\\& & 9& 15& \color{black}{45} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{15}&\color{orangered}{52} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-4x^{2}+7 }{ x-3 } = \color{blue}{3x^{2}+5x+15} ~+~ \frac{ \color{red}{ 52 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-4&0&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&-4&0&7\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-4&0&7\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 9 } = \color{orangered}{ 5 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ -4 }&0&7\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{5}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 5 } = \color{blue}{ 15 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-4&0&7\\& & 9& \color{blue}{15} & \\ \hline &3&\color{blue}{5}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 15 } = \color{orangered}{ 15 } $
$$ \begin{array}{c|rrrr}3&3&-4&\color{orangered}{ 0 }&7\\& & 9& \color{orangered}{15} & \\ \hline &3&5&\color{orangered}{15}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-4&0&7\\& & 9& 15& \color{blue}{45} \\ \hline &3&5&\color{blue}{15}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 45 } = \color{orangered}{ 52 } $
$$ \begin{array}{c|rrrr}3&3&-4&0&\color{orangered}{ 7 }\\& & 9& 15& \color{orangered}{45} \\ \hline &\color{blue}{3}&\color{blue}{5}&\color{blue}{15}&\color{orangered}{52} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+5x+15 } $ with a remainder of $ \color{red}{ 52 } $.