Divide using synthetic division $$ ( x^{3}+4x^{2}-19x+14 ) \div ( 0 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&1&4&-19&14\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-19}&\color{orangered}{14} \end{array} $$The solution is:
$$ \frac{ x^{3}+4x^{2}-19x+14 }{ x } = \color{blue}{x^{2}+4x-19} ~+~ \frac{ \color{red}{ 14 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&4&-19&14\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 1 }&4&-19&14\\& & & & \\ \hline &\color{orangered}{1}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&4&-19&14\\& & \color{blue}{0} & & \\ \hline &\color{blue}{1}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 4 } + \color{orangered}{ 0 } = \color{orangered}{ 4 } $
$$ \begin{array}{c|rrrr}0&1&\color{orangered}{ 4 }&-19&14\\& & \color{orangered}{0} & & \\ \hline &1&\color{orangered}{4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 4 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&4&-19&14\\& & 0& \color{blue}{0} & \\ \hline &1&\color{blue}{4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 0 } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}0&1&4&\color{orangered}{ -19 }&14\\& & 0& \color{orangered}{0} & \\ \hline &1&4&\color{orangered}{-19}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&1&4&-19&14\\& & 0& 0& \color{blue}{0} \\ \hline &1&4&\color{blue}{-19}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 14 } + \color{orangered}{ 0 } = \color{orangered}{ 14 } $
$$ \begin{array}{c|rrrr}0&1&4&-19&\color{orangered}{ 14 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{4}&\color{blue}{-19}&\color{orangered}{14} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{2}+4x-19 } $ with a remainder of $ \color{red}{ 14 } $.