Divide using synthetic division $$ ( 15x^{3}+34x^{2}-31x+6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&15&34&-31&6\\& & 45& 237& \color{black}{618} \\ \hline &\color{blue}{15}&\color{blue}{79}&\color{blue}{206}&\color{orangered}{624} \end{array} $$The solution is:
$$ \frac{ 15x^{3}+34x^{2}-31x+6 }{ x-3 } = \color{blue}{15x^{2}+79x+206} ~+~ \frac{ \color{red}{ 624 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&15&34&-31&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 15 }&34&-31&6\\& & & & \\ \hline &\color{orangered}{15}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 15 } = \color{blue}{ 45 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&15&34&-31&6\\& & \color{blue}{45} & & \\ \hline &\color{blue}{15}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 34 } + \color{orangered}{ 45 } = \color{orangered}{ 79 } $
$$ \begin{array}{c|rrrr}3&15&\color{orangered}{ 34 }&-31&6\\& & \color{orangered}{45} & & \\ \hline &15&\color{orangered}{79}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 79 } = \color{blue}{ 237 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&15&34&-31&6\\& & 45& \color{blue}{237} & \\ \hline &15&\color{blue}{79}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -31 } + \color{orangered}{ 237 } = \color{orangered}{ 206 } $
$$ \begin{array}{c|rrrr}3&15&34&\color{orangered}{ -31 }&6\\& & 45& \color{orangered}{237} & \\ \hline &15&79&\color{orangered}{206}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 206 } = \color{blue}{ 618 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&15&34&-31&6\\& & 45& 237& \color{blue}{618} \\ \hline &15&79&\color{blue}{206}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 618 } = \color{orangered}{ 624 } $
$$ \begin{array}{c|rrrr}3&15&34&-31&\color{orangered}{ 6 }\\& & 45& 237& \color{orangered}{618} \\ \hline &\color{blue}{15}&\color{blue}{79}&\color{blue}{206}&\color{orangered}{624} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 15x^{2}+79x+206 } $ with a remainder of $ \color{red}{ 624 } $.