Divide using synthetic division $$ ( 3x^{3}+4x^{2}-7x+6 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&3&4&-7&6\\& & 9& 39& \color{black}{96} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{32}&\color{orangered}{102} \end{array} $$The solution is:
$$ \frac{ 3x^{3}+4x^{2}-7x+6 }{ x-3 } = \color{blue}{3x^{2}+13x+32} ~+~ \frac{ \color{red}{ 102 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&4&-7&6\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 3 }&4&-7&6\\& & & & \\ \hline &\color{orangered}{3}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 3 } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&4&-7&6\\& & \color{blue}{9} & & \\ \hline &\color{blue}{3}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 9 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}3&3&\color{orangered}{ 4 }&-7&6\\& & \color{orangered}{9} & & \\ \hline &3&\color{orangered}{13}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 13 } = \color{blue}{ 39 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&4&-7&6\\& & 9& \color{blue}{39} & \\ \hline &3&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 39 } = \color{orangered}{ 32 } $
$$ \begin{array}{c|rrrr}3&3&4&\color{orangered}{ -7 }&6\\& & 9& \color{orangered}{39} & \\ \hline &3&13&\color{orangered}{32}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 32 } = \color{blue}{ 96 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&3&4&-7&6\\& & 9& 39& \color{blue}{96} \\ \hline &3&13&\color{blue}{32}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 6 } + \color{orangered}{ 96 } = \color{orangered}{ 102 } $
$$ \begin{array}{c|rrrr}3&3&4&-7&\color{orangered}{ 6 }\\& & 9& 39& \color{orangered}{96} \\ \hline &\color{blue}{3}&\color{blue}{13}&\color{blue}{32}&\color{orangered}{102} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 3x^{2}+13x+32 } $ with a remainder of $ \color{red}{ 102 } $.