Divide using synthetic division $$ ( 2x^{3}+7x^{2}-5x+4 ) \div ( x-3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}3&2&7&-5&4\\& & 6& 39& \color{black}{102} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{34}&\color{orangered}{106} \end{array} $$The solution is:
$$ \frac{ 2x^{3}+7x^{2}-5x+4 }{ x-3 } = \color{blue}{2x^{2}+13x+34} ~+~ \frac{ \color{red}{ 106 } }{ x-3 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&7&-5&4\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}3&\color{orangered}{ 2 }&7&-5&4\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 2 } = \color{blue}{ 6 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&7&-5&4\\& & \color{blue}{6} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 6 } = \color{orangered}{ 13 } $
$$ \begin{array}{c|rrrr}3&2&\color{orangered}{ 7 }&-5&4\\& & \color{orangered}{6} & & \\ \hline &2&\color{orangered}{13}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&7&-5&4\\& & 6& \color{blue}{39} & \\ \hline &2&\color{blue}{13}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -5 } + \color{orangered}{ 39 } = \color{orangered}{ 34 } $
$$ \begin{array}{c|rrrr}3&2&7&\color{orangered}{ -5 }&4\\& & 6& \color{orangered}{39} & \\ \hline &2&13&\color{orangered}{34}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 3 } \cdot \color{blue}{ 34 } = \color{blue}{ 102 } $.
$$ \begin{array}{c|rrrr}\color{blue}{3}&2&7&-5&4\\& & 6& 39& \color{blue}{102} \\ \hline &2&13&\color{blue}{34}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 102 } = \color{orangered}{ 106 } $
$$ \begin{array}{c|rrrr}3&2&7&-5&\color{orangered}{ 4 }\\& & 6& 39& \color{orangered}{102} \\ \hline &\color{blue}{2}&\color{blue}{13}&\color{blue}{34}&\color{orangered}{106} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}+13x+34 } $ with a remainder of $ \color{red}{ 106 } $.