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