Divide using synthetic division $$ ( 14x^{3}-9x^{2} ) \div ( 5x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&14&-9&0&0\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{14}&\color{blue}{-9}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 14x^{3}-9x^{2} }{ x } = \color{blue}{14x^{2}-9x} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&14&-9&0&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 14 }&-9&0&0\\& & & & \\ \hline &\color{orangered}{14}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 14 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&14&-9&0&0\\& & \color{blue}{0} & & \\ \hline &\color{blue}{14}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -9 } + \color{orangered}{ 0 } = \color{orangered}{ -9 } $
$$ \begin{array}{c|rrrr}0&14&\color{orangered}{ -9 }&0&0\\& & \color{orangered}{0} & & \\ \hline &14&\color{orangered}{-9}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -9 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&14&-9&0&0\\& & 0& \color{blue}{0} & \\ \hline &14&\color{blue}{-9}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&14&-9&\color{orangered}{ 0 }&0\\& & 0& \color{orangered}{0} & \\ \hline &14&-9&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&14&-9&0&0\\& & 0& 0& \color{blue}{0} \\ \hline &14&-9&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&14&-9&0&\color{orangered}{ 0 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{14}&\color{blue}{-9}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 14x^{2}-9x } $ with a remainder of $ \color{red}{ 0 } $.