Divide using synthetic division $$ ( 2x^{3}-4x^{2}-14x+28 ) \div ( 1 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&2&-4&-14&28\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{-14}&\color{orangered}{28} \end{array} $$The solution is:
$$ \frac{ 2x^{3}-4x^{2}-14x+28 }{ x } = \color{blue}{2x^{2}-4x-14} ~+~ \frac{ \color{red}{ 28 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-4&-14&28\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 2 }&-4&-14&28\\& & & & \\ \hline &\color{orangered}{2}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-4&-14&28\\& & \color{blue}{0} & & \\ \hline &\color{blue}{2}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -4 } + \color{orangered}{ 0 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrr}0&2&\color{orangered}{ -4 }&-14&28\\& & \color{orangered}{0} & & \\ \hline &2&\color{orangered}{-4}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-4&-14&28\\& & 0& \color{blue}{0} & \\ \hline &2&\color{blue}{-4}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -14 } + \color{orangered}{ 0 } = \color{orangered}{ -14 } $
$$ \begin{array}{c|rrrr}0&2&-4&\color{orangered}{ -14 }&28\\& & 0& \color{orangered}{0} & \\ \hline &2&-4&\color{orangered}{-14}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -14 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&2&-4&-14&28\\& & 0& 0& \color{blue}{0} \\ \hline &2&-4&\color{blue}{-14}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 28 } + \color{orangered}{ 0 } = \color{orangered}{ 28 } $
$$ \begin{array}{c|rrrr}0&2&-4&-14&\color{orangered}{ 28 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{2}&\color{blue}{-4}&\color{blue}{-14}&\color{orangered}{28} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 2x^{2}-4x-14 } $ with a remainder of $ \color{red}{ 28 } $.