Divide using synthetic division $$ ( 5x^{4}+12x^{3}-21x^{2}-40x-12 ) \div ( 3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}0&5&12&-21&-40&-12\\& & 0& 0& 0& \color{black}{0} \\ \hline &\color{blue}{5}&\color{blue}{12}&\color{blue}{-21}&\color{blue}{-40}&\color{orangered}{-12} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+12x^{3}-21x^{2}-40x-12 }{ x } = \color{blue}{5x^{3}+12x^{2}-21x-40} \color{red}{~-~} \frac{ \color{red}{ 12 } }{ x } $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&12&-21&-40&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}0&\color{orangered}{ 5 }&12&-21&-40&-12\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 5 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&12&-21&-40&-12\\& & \color{blue}{0} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ 0 } = \color{orangered}{ 12 } $
$$ \begin{array}{c|rrrrr}0&5&\color{orangered}{ 12 }&-21&-40&-12\\& & \color{orangered}{0} & & & \\ \hline &5&\color{orangered}{12}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 12 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&12&-21&-40&-12\\& & 0& \color{blue}{0} & & \\ \hline &5&\color{blue}{12}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 0 } = \color{orangered}{ -21 } $
$$ \begin{array}{c|rrrrr}0&5&12&\color{orangered}{ -21 }&-40&-12\\& & 0& \color{orangered}{0} & & \\ \hline &5&12&\color{orangered}{-21}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -21 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&12&-21&-40&-12\\& & 0& 0& \color{blue}{0} & \\ \hline &5&12&\color{blue}{-21}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 0 } = \color{orangered}{ -40 } $
$$ \begin{array}{c|rrrrr}0&5&12&-21&\color{orangered}{ -40 }&-12\\& & 0& 0& \color{orangered}{0} & \\ \hline &5&12&-21&\color{orangered}{-40}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -40 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{0}&5&12&-21&-40&-12\\& & 0& 0& 0& \color{blue}{0} \\ \hline &5&12&-21&\color{blue}{-40}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 0 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}0&5&12&-21&-40&\color{orangered}{ -12 }\\& & 0& 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{5}&\color{blue}{12}&\color{blue}{-21}&\color{blue}{-40}&\color{orangered}{-12} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}+12x^{2}-21x-40 } $ with a remainder of $ \color{red}{ -12 } $.