Divide using synthetic division $$ ( 3x^{3}-6x^{2}+x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&-6&1&0\\& & 9& 9& \color{black}{30} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{30} \end{array} $$The solution is:
$$ \frac{ 3x^{3}-6x^{2}+x }{ x-3 } = \color{blue}{3x^{2}+3x+10} ~+~ \frac{ \color{red}{ 30 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-6&1&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&-6&1&0\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-6&1&0\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -6 } + \color{orangered}{ 9 } = \color{orangered}{ 3 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ -6 }&1&0\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{3}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-6&1&0\\& & 9& \color{blue}{9} & \\ \hline &3&\color{blue}{3}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 9 } = \color{orangered}{ 10 } $
$$ \begin{array}{c|rrrr}3&3&-6&\color{orangered}{ 1 }&0\\& & 9& \color{orangered}{9} & \\ \hline &3&3&\color{orangered}{10}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 10 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&-6&1&0\\& & 9& 9& \color{blue}{30} \\ \hline &3&3&\color{blue}{10}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 30 } = \color{orangered}{ 30 } $
$$ \begin{array}{c|rrrr}3&3&-6&1&\color{orangered}{ 0 }\\& & 9& 9& \color{orangered}{30} \\ \hline &\color{blue}{3}&\color{blue}{3}&\color{blue}{10}&\color{orangered}{30} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+3x+10 } $ with a remainder of $ \color{red}{ 30 } $.