Divide using synthetic division $$ ( x^{4}+2x^{3}+x^{2}+8x-12 ) \div ( -3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&1&2&1&8&-12\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{1}&\color{blue}{8}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ x^{4}+2x^{3}+x^{2}+8x-12 }{ x } = \color{blue}{x^{3}+2x^{2}+x+8} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&2&1&8&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 1 }&2&1&8&-12\\& & & & & \\ \hline &\color{orangered}{1}&&&& \end{array} $$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 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&2&1&8&-12\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{1}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 2 } + \color{orangered}{ 0 } = \color{orangered}{ 2 } $
$$ \begin{array}{c|rrrrr}0&1&\color{orangered}{ 2 }&1&8&-12\\& & \color{orangered}{0} & & & \\ \hline &1&\color{orangered}{2}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 2 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&2&1&8&-12\\& & 0& \color{blue}{0} & & \\ \hline &1&\color{blue}{2}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 1 } + \color{orangered}{ 0 } = \color{orangered}{ 1 } $
$$ \begin{array}{c|rrrrr}0&1&2&\color{orangered}{ 1 }&8&-12\\& & 0& \color{orangered}{0} & & \\ \hline &1&2&\color{orangered}{1}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 1 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&2&1&8&-12\\& & 0& 0& \color{blue}{0} & \\ \hline &1&2&\color{blue}{1}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 8 } + \color{orangered}{ 0 } = \color{orangered}{ 8 } $
$$ \begin{array}{c|rrrrr}0&1&2&1&\color{orangered}{ 8 }&-12\\& & 0& 0& \color{orangered}{0} & \\ \hline &1&2&1&\color{orangered}{8}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 8 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&1&2&1&8&-12\\& & 0& 0& 0& \color{blue}{0} \\ \hline &1&2&1&\color{blue}{8}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}0&1&2&1&8&\color{orangered}{ -12 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{1}&\color{blue}{2}&\color{blue}{1}&\color{blue}{8}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ x^{3}+2x^{2}+x+8 } $ with a remainder of $ \color{red}{ -12 } $.