Divide using synthetic division $$ ( 5x^{4}+12x^{3}-21x^{2}-40x-12 ) \div ( x+3 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrrr}-3&5&12&-21&-40&-12\\& & -15& 9& 36& \color{black}{12} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-12}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 5x^{4}+12x^{3}-21x^{2}-40x-12 }{ x+3 } = \color{blue}{5x^{3}-3x^{2}-12x-4} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table. Put the zero from $ x + 3 = 0 $ ( $ x = \color{blue}{ -3 } $ ) at the left.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&12&-21&-40&-12\\& & & & & \\ \hline &&&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrrr}-3&\color{orangered}{ 5 }&12&-21&-40&-12\\& & & & & \\ \hline &\color{orangered}{5}&&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ 5 } = \color{blue}{ -15 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&12&-21&-40&-12\\& & \color{blue}{-15} & & & \\ \hline &\color{blue}{5}&&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ 12 } + \color{orangered}{ \left( -15 \right) } = \color{orangered}{ -3 } $
$$ \begin{array}{c|rrrrr}-3&5&\color{orangered}{ 12 }&-21&-40&-12\\& & \color{orangered}{-15} & & & \\ \hline &5&\color{orangered}{-3}&&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -3 \right) } = \color{blue}{ 9 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&12&-21&-40&-12\\& & -15& \color{blue}{9} & & \\ \hline &5&\color{blue}{-3}&&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ -21 } + \color{orangered}{ 9 } = \color{orangered}{ -12 } $
$$ \begin{array}{c|rrrrr}-3&5&12&\color{orangered}{ -21 }&-40&-12\\& & -15& \color{orangered}{9} & & \\ \hline &5&-3&\color{orangered}{-12}&& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -12 \right) } = \color{blue}{ 36 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&12&-21&-40&-12\\& & -15& 9& \color{blue}{36} & \\ \hline &5&-3&\color{blue}{-12}&& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ -40 } + \color{orangered}{ 36 } = \color{orangered}{ -4 } $
$$ \begin{array}{c|rrrrr}-3&5&12&-21&\color{orangered}{ -40 }&-12\\& & -15& 9& \color{orangered}{36} & \\ \hline &5&-3&-12&\color{orangered}{-4}& \end{array} $$Step 8 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ -3 } \cdot \color{blue}{ \left( -4 \right) } = \color{blue}{ 12 } $.
$$ \begin{array}{c|rrrrr}\color{blue}{-3}&5&12&-21&-40&-12\\& & -15& 9& 36& \color{blue}{12} \\ \hline &5&-3&-12&\color{blue}{-4}& \end{array} $$Step 9 : Add down last column: $ \color{orangered}{ -12 } + \color{orangered}{ 12 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrrr}-3&5&12&-21&-40&\color{orangered}{ -12 }\\& & -15& 9& 36& \color{orangered}{12} \\ \hline &\color{blue}{5}&\color{blue}{-3}&\color{blue}{-12}&\color{blue}{-4}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 5x^{3}-3x^{2}-12x-4 } $ with a remainder of $ \color{red}{ 0 } $.