Divide using synthetic division $$ ( 6x^{3}+4x^{2}+3 ) \div ( 6x-4 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}4&6&4&0&3\\& & 24& 112& \color{black}{448} \\ \hline &\color{blue}{6}&\color{blue}{28}&\color{blue}{112}&\color{orangered}{451} \end{array} $$The solution is:
$$ \frac{ 6x^{3}+4x^{2}+3 }{ x-4 } = \color{blue}{6x^{2}+28x+112} ~+~ \frac{ \color{red}{ 451 } }{ x-4 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -4 = 0 $ ( $ x = \color{blue}{ 4 } $ ) at the left.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&4&0&3\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}4&\color{orangered}{ 6 }&4&0&3\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 6 } = \color{blue}{ 24 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&4&0&3\\& & \color{blue}{24} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 24 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}4&6&\color{orangered}{ 4 }&0&3\\& & \color{orangered}{24} & & \\ \hline &6&\color{orangered}{28}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 28 } = \color{blue}{ 112 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&4&0&3\\& & 24& \color{blue}{112} & \\ \hline &6&\color{blue}{28}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 112 } = \color{orangered}{ 112 } $
$$ \begin{array}{c|rrrr}4&6&4&\color{orangered}{ 0 }&3\\& & 24& \color{orangered}{112} & \\ \hline &6&28&\color{orangered}{112}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 4 } \cdot \color{blue}{ 112 } = \color{blue}{ 448 } $.
$$ \begin{array}{c|rrrr}\color{blue}{4}&6&4&0&3\\& & 24& 112& \color{blue}{448} \\ \hline &6&28&\color{blue}{112}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 3 } + \color{orangered}{ 448 } = \color{orangered}{ 451 } $
$$ \begin{array}{c|rrrr}4&6&4&0&\color{orangered}{ 3 }\\& & 24& 112& \color{orangered}{448} \\ \hline &\color{blue}{6}&\color{blue}{28}&\color{blue}{112}&\color{orangered}{451} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+28x+112 } $ with a remainder of $ \color{red}{ 451 } $.