Divide using synthetic division $$ ( x^{3}+x^{2}+9x ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&1&9&0\\& & 3& 12& \color{black}{63} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{63} \end{array} $$The solution is:
$$ \frac{ x^{3}+x^{2}+9x }{ x-3 } = \color{blue}{x^{2}+4x+21} ~+~ \frac{ \color{red}{ 63 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&9&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&1&9&0\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 1 } = \color{blue}{ 3 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&9&0\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 3 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 1 }&9&0\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 4 } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&9&0\\& & 3& \color{blue}{12} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 9 } + \color{orangered}{ 12 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}3&1&1&\color{orangered}{ 9 }&0\\& & 3& \color{orangered}{12} & \\ \hline &1&4&\color{orangered}{21}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 21 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&1&9&0\\& & 3& 12& \color{blue}{63} \\ \hline &1&4&\color{blue}{21}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 63 } = \color{orangered}{ 63 } $
$$ \begin{array}{c|rrrr}3&1&1&9&\color{orangered}{ 0 }\\& & 3& 12& \color{orangered}{63} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{21}&\color{orangered}{63} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+4x+21 } $ with a remainder of $ \color{red}{ 63 } $.