Divide using synthetic division $$ ( 6x^{2}-7x+14 ) \div ( 2x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrr}5&6&-7&14\\& & 30& \color{black}{115} \\ \hline &\color{blue}{6}&\color{blue}{23}&\color{orangered}{129} \end{array} $$The solution is:
$$ \frac{ 6x^{2}-7x+14 }{ x-5 } = \color{blue}{6x+23} ~+~ \frac{ \color{red}{ 129 } }{ x-5 } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x -5 = 0 $ ( $ x = \color{blue}{ 5 } $ ) at the left.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&-7&14\\& & & \\ \hline &&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrr}5&\color{orangered}{ 6 }&-7&14\\& & & \\ \hline &\color{orangered}{6}&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 6 } = \color{blue}{ 30 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&-7&14\\& & \color{blue}{30} & \\ \hline &\color{blue}{6}&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 30 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrr}5&6&\color{orangered}{ -7 }&14\\& & \color{orangered}{30} & \\ \hline &6&\color{orangered}{23}& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 23 } = \color{blue}{ 115 } $.
$$ \begin{array}{c|rrr}\color{blue}{5}&6&-7&14\\& & 30& \color{blue}{115} \\ \hline &6&\color{blue}{23}& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 115 } = \color{orangered}{ 129 } $
$$ \begin{array}{c|rrr}5&6&-7&\color{orangered}{ 14 }\\& & 30& \color{orangered}{115} \\ \hline &\color{blue}{6}&\color{blue}{23}&\color{orangered}{129} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x+23 } $ with a remainder of $ \color{red}{ 129 } $.