Divide using synthetic division $$ ( x^{2}+6x+5 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}3&1&6&5\\& & 3& \color{black}{27} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{orangered}{32} \end{array} $$The solution is:
$$ \frac{ x^{2}+6x+5 }{ x-3 } = \color{blue}{x+9} ~+~ \frac{ \color{red}{ 32 } }{ x-3 } $$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.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&6&5\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}3&\color{orangered}{ 1 }&6&5\\& & & \\ \hline &\color{orangered}{1}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&6&5\\& & \color{blue}{3} & \\ \hline &\color{blue}{1}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 3 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrr}3&1&\color{orangered}{ 6 }&5\\& & \color{orangered}{3} & \\ \hline &1&\color{orangered}{9}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 9 } = \color{blue}{ 27 } $.
$$ \begin{array}{c|rrr}\color{blue}{3}&1&6&5\\& & 3& \color{blue}{27} \\ \hline &1&\color{blue}{9}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 5 } + \color{orangered}{ 27 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrr}3&1&6&\color{orangered}{ 5 }\\& & 3& \color{orangered}{27} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{orangered}{32} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x+9 } $ with a remainder of $ \color{red}{ 32 } $.