Divide using synthetic division $$ ( 6x^{3}-7x^{2}-22x+7 ) \div ( 2x-5 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}5&6&-7&-22&7\\& & 30& 115& \color{black}{465} \\ \hline &\color{blue}{6}&\color{blue}{23}&\color{blue}{93}&\color{orangered}{472} \end{array} $$The solution is:
$$ \frac{ 6x^{3}-7x^{2}-22x+7 }{ x-5 } = \color{blue}{6x^{2}+23x+93} ~+~ \frac{ \color{red}{ 472 } }{ x-5 } $$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.
$$ \begin{array}{c|rrrr}\color{blue}{5}&6&-7&-22&7\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}5&\color{orangered}{ 6 }&-7&-22&7\\& & & & \\ \hline &\color{orangered}{6}&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&6&-7&-22&7\\& & \color{blue}{30} & & \\ \hline &\color{blue}{6}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 30 } = \color{orangered}{ 23 } $
$$ \begin{array}{c|rrrr}5&6&\color{orangered}{ -7 }&-22&7\\& & \color{orangered}{30} & & \\ \hline &6&\color{orangered}{23}&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&6&-7&-22&7\\& & 30& \color{blue}{115} & \\ \hline &6&\color{blue}{23}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -22 } + \color{orangered}{ 115 } = \color{orangered}{ 93 } $
$$ \begin{array}{c|rrrr}5&6&-7&\color{orangered}{ -22 }&7\\& & 30& \color{orangered}{115} & \\ \hline &6&23&\color{orangered}{93}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 5 } \cdot \color{blue}{ 93 } = \color{blue}{ 465 } $.
$$ \begin{array}{c|rrrr}\color{blue}{5}&6&-7&-22&7\\& & 30& 115& \color{blue}{465} \\ \hline &6&23&\color{blue}{93}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 7 } + \color{orangered}{ 465 } = \color{orangered}{ 472 } $
$$ \begin{array}{c|rrrr}5&6&-7&-22&\color{orangered}{ 7 }\\& & 30& 115& \color{orangered}{465} \\ \hline &\color{blue}{6}&\color{blue}{23}&\color{blue}{93}&\color{orangered}{472} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 6x^{2}+23x+93 } $ with a remainder of $ \color{red}{ 472 } $.