Divide using synthetic division $$ ( 12x^{3}-x^{2}-7x+2 ) \div ( x ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&12&-1&-7&2\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{12}&\color{blue}{-1}&\color{blue}{-7}&\color{orangered}{2} \end{array} $$The solution is:
$$ \frac{ 12x^{3}-x^{2}-7x+2 }{ x } = \color{blue}{12x^{2}-x-7} ~+~ \frac{ \color{red}{ 2 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&12&-1&-7&2\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 12 }&-1&-7&2\\& & & & \\ \hline &\color{orangered}{12}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&12&-1&-7&2\\& & \color{blue}{0} & & \\ \hline &\color{blue}{12}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -1 } + \color{orangered}{ 0 } = \color{orangered}{ -1 } $
$$ \begin{array}{c|rrrr}0&12&\color{orangered}{ -1 }&-7&2\\& & \color{orangered}{0} & & \\ \hline &12&\color{orangered}{-1}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -1 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&12&-1&-7&2\\& & 0& \color{blue}{0} & \\ \hline &12&\color{blue}{-1}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -7 } + \color{orangered}{ 0 } = \color{orangered}{ -7 } $
$$ \begin{array}{c|rrrr}0&12&-1&\color{orangered}{ -7 }&2\\& & 0& \color{orangered}{0} & \\ \hline &12&-1&\color{orangered}{-7}& \end{array} $$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 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&12&-1&-7&2\\& & 0& 0& \color{blue}{0} \\ \hline &12&-1&\color{blue}{-7}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrr}0&12&-1&-7&\color{orangered}{ 2 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{12}&\color{blue}{-1}&\color{blue}{-7}&\color{orangered}{2} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 12x^{2}-x-7 } $ with a remainder of $ \color{red}{ 2 } $.