Divide using synthetic division $$ ( x^{3}+6x^{2}+10x-3 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&1&6&10&-3\\& & 3& 27& \color{black}{111} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{37}&\color{orangered}{108} \end{array} $$The solution is:
$$ \frac{ x^{3}+6x^{2}+10x-3 }{ x-3 } = \color{blue}{x^{2}+9x+37} ~+~ \frac{ \color{red}{ 108 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&10&-3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 1 }&6&10&-3\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&10&-3\\& & \color{blue}{3} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 3 } = \color{orangered}{ 9 } $
$$ \begin{array}{c|rrrr}3&1&\color{orangered}{ 6 }&10&-3\\& & \color{orangered}{3} & & \\ \hline &1&\color{orangered}{9}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&10&-3\\& & 3& \color{blue}{27} & \\ \hline &1&\color{blue}{9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 27 } = \color{orangered}{ 37 } $
$$ \begin{array}{c|rrrr}3&1&6&\color{orangered}{ 10 }&-3\\& & 3& \color{orangered}{27} & \\ \hline &1&9&\color{orangered}{37}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 37 } = \color{blue}{ 111 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&1&6&10&-3\\& & 3& 27& \color{blue}{111} \\ \hline &1&9&\color{blue}{37}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -3 } + \color{orangered}{ 111 } = \color{orangered}{ 108 } $
$$ \begin{array}{c|rrrr}3&1&6&10&\color{orangered}{ -3 }\\& & 3& 27& \color{orangered}{111} \\ \hline &\color{blue}{1}&\color{blue}{9}&\color{blue}{37}&\color{orangered}{108} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+9x+37 } $ with a remainder of $ \color{red}{ 108 } $.