Divide using synthetic division $$ ( 6x^{3}+3x^{2}+10x+14 ) \div ( 2x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&6&3&10&14\\& & 18& 63& \color{black}{219} \\ \hline &\color{blue}{6}&\color{blue}{21}&\color{blue}{73}&\color{orangered}{233} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+3x^{2}+10x+14 }{ x-3 } = \color{blue}{6x^{2}+21x+73} ~+~ \frac{ \color{red}{ 233 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&3&10&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 6 }&3&10&14\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 6 } = \color{blue}{ 18 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&3&10&14\\& & \color{blue}{18} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 18 } = \color{orangered}{ 21 } $
$$ \begin{array}{c|rrrr}3&6&\color{orangered}{ 3 }&10&14\\& & \color{orangered}{18} & & \\ \hline &6&\color{orangered}{21}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 21 } = \color{blue}{ 63 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&3&10&14\\& & 18& \color{blue}{63} & \\ \hline &6&\color{blue}{21}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 10 } + \color{orangered}{ 63 } = \color{orangered}{ 73 } $
$$ \begin{array}{c|rrrr}3&6&3&\color{orangered}{ 10 }&14\\& & 18& \color{orangered}{63} & \\ \hline &6&21&\color{orangered}{73}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 73 } = \color{blue}{ 219 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&6&3&10&14\\& & 18& 63& \color{blue}{219} \\ \hline &6&21&\color{blue}{73}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 219 } = \color{orangered}{ 233 } $
$$ \begin{array}{c|rrrr}3&6&3&10&\color{orangered}{ 14 }\\& & 18& 63& \color{orangered}{219} \\ \hline &\color{blue}{6}&\color{blue}{21}&\color{blue}{73}&\color{orangered}{233} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+21x+73 } $ with a remainder of $ \color{red}{ 233 } $.