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